EXAMPLES
Coordinate Geometry • 10 Questions
Question 1
Hint available
Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.
Key Idea
Use the slope (or area) test to check whether three points are collinear. If they are not collinear, they form a triangle. Then apply the distance formula to find the lengths of the sides and use the Pythagoras theorem to identify the type of triangle (right‑angled, isosceles, equilateral, etc.).
Step-by-Step Solution
1. Check collinearity
- Let the points be \(A(3,2)\), \(B(-2,-3)\) and \(C(2,3)\).
- Slope of \(AB\):
$$m_{AB}=\frac{-3-2}{-2-3}=\frac{-5}{-5}=1$$
- Slope of \(AC\):
$$m_{AC}=\frac{3-2}{2-3}=\frac{1}{-1}=-1$$
- Since \(m_{AB}
eq m_{AC}\), the points are not collinear. Hence they form a triangle.
2. Find the lengths of the sides using the distance formula
- \(AB\):
$$AB=\sqrt{(3-(-2))^{2}+(2-(-3))^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}$$
- \(BC\):
$$BC=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16+36}=\sqrt{52}$$
- \(AC\):
$$AC=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$$
3. Identify the type of triangle
- Compute the squares of the sides: \(AB^{2}=50\), \(BC^{2}=52\), \(AC^{2}=2\).
- The largest side is \(BC\). Check Pythagoras theorem: \(AB^{2}+AC^{2}=50+2=52=BC^{2}\).
- Since the sum of the squares of the two smaller sides equals the square of the largest side, the triangle satisfies the condition for a right‑angled triangle.
- The right angle is at point \(A(3,2)\) because the sides meeting at \(A\) are \(AB\) and \(AC\).
4. Conclusion
- The three points do form a triangle, and the triangle is a right‑angled triangle (not isosceles or equilateral).
- Let the points be \(A(3,2)\), \(B(-2,-3)\) and \(C(2,3)\).
- Slope of \(AB\):
$$m_{AB}=\frac{-3-2}{-2-3}=\frac{-5}{-5}=1$$
- Slope of \(AC\):
$$m_{AC}=\frac{3-2}{2-3}=\frac{1}{-1}=-1$$
- Since \(m_{AB}
eq m_{AC}\), the points are not collinear. Hence they form a triangle.
2. Find the lengths of the sides using the distance formula
- \(AB\):
$$AB=\sqrt{(3-(-2))^{2}+(2-(-3))^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}$$
- \(BC\):
$$BC=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16+36}=\sqrt{52}$$
- \(AC\):
$$AC=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$$
3. Identify the type of triangle
- Compute the squares of the sides: \(AB^{2}=50\), \(BC^{2}=52\), \(AC^{2}=2\).
- The largest side is \(BC\). Check Pythagoras theorem: \(AB^{2}+AC^{2}=50+2=52=BC^{2}\).
- Since the sum of the squares of the two smaller sides equals the square of the largest side, the triangle satisfies the condition for a right‑angled triangle.
- The right angle is at point \(A(3,2)\) because the sides meeting at \(A\) are \(AB\) and \(AC\).
4. Conclusion
- The three points do form a triangle, and the triangle is a right‑angled triangle (not isosceles or equilateral).
Question 2
Hint available
Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Key Idea
In a square all four sides are equal and the two diagonals are equal. Using the distance formula we can compute the lengths of all possible line segments joining the given points. If we can arrange the points so that four equal sides and two equal diagonals are obtained, the quadrilateral is a square. Additionally, the product of slopes of adjacent sides must be –1 (perpendicular).
Step-by-Step Solution
1. List the points:\
\(A(1,7),\; B(4,2),\; C(-1,-1),\; D(-4,4)\).
2. Compute distances between every pair using the distance formula \(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
- \(AB = \sqrt{(4-1)^2+(2-7)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}\)
- \(AD = \sqrt{(-4-1)^2+(4-7)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\)
- \(BC = \sqrt{(-1-4)^2+(-1-2)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{34}\)
- \(CD = \sqrt{(-4+1)^2+(4+1)^2}=\sqrt{(-3)^2+5^2}=\sqrt{34}\)
- \(AC = \sqrt{(-1-1)^2+(-1-7)^2}=\sqrt{(-2)^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}\)
- \(BD = \sqrt{(-4-4)^2+(4-2)^2}=\sqrt{(-8)^2+2^2}=\sqrt{64+4}=\sqrt{68}\)
3. Identify the ordering: Since \(AB, BC, CD, DA\) are all \(\sqrt{34}\) and the two remaining distances \(AC\) and \(BD\) are \(\sqrt{68}\), the points can be taken in the order \(A\to B\to C\to D\to A\).
4. Check perpendicularity (optional but confirms square):
- Slope of \(AB\): \(m_{AB}=\frac{2-7}{4-1}=\frac{-5}{3}\).
- Slope of \(BC\): \(m_{BC}=\frac{-1-2}{-1-4}=\frac{-3}{-5}=\frac{3}{5}\).
- \(m_{AB}\times m_{BC}=\frac{-5}{3}\times\frac{3}{5}=-1\). Hence \(AB\perp BC\).
- Similarly \(m_{CD}=\frac{4+1}{-4+1}=\frac{5}{-3}= -\frac{5}{3}\) and \(m_{DA}=\frac{7-4}{1+4}=\frac{3}{5}\); their product is also \(-1\).
5. Conclusion: All four sides are equal (\(\sqrt{34}\)), the two diagonals are equal (\(\sqrt{68}\)), and adjacent sides are perpendicular. Therefore the quadrilateral formed by the given points is a square.
6. Answer: The points \((1,7), (4,2), (-1,-1), (-4,4)\) are indeed the vertices of a square.
\(A(1,7),\; B(4,2),\; C(-1,-1),\; D(-4,4)\).
2. Compute distances between every pair using the distance formula \(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
- \(AB = \sqrt{(4-1)^2+(2-7)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}\)
- \(AD = \sqrt{(-4-1)^2+(4-7)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\)
- \(BC = \sqrt{(-1-4)^2+(-1-2)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{34}\)
- \(CD = \sqrt{(-4+1)^2+(4+1)^2}=\sqrt{(-3)^2+5^2}=\sqrt{34}\)
- \(AC = \sqrt{(-1-1)^2+(-1-7)^2}=\sqrt{(-2)^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}\)
- \(BD = \sqrt{(-4-4)^2+(4-2)^2}=\sqrt{(-8)^2+2^2}=\sqrt{64+4}=\sqrt{68}\)
3. Identify the ordering: Since \(AB, BC, CD, DA\) are all \(\sqrt{34}\) and the two remaining distances \(AC\) and \(BD\) are \(\sqrt{68}\), the points can be taken in the order \(A\to B\to C\to D\to A\).
4. Check perpendicularity (optional but confirms square):
- Slope of \(AB\): \(m_{AB}=\frac{2-7}{4-1}=\frac{-5}{3}\).
- Slope of \(BC\): \(m_{BC}=\frac{-1-2}{-1-4}=\frac{-3}{-5}=\frac{3}{5}\).
- \(m_{AB}\times m_{BC}=\frac{-5}{3}\times\frac{3}{5}=-1\). Hence \(AB\perp BC\).
- Similarly \(m_{CD}=\frac{4+1}{-4+1}=\frac{5}{-3}= -\frac{5}{3}\) and \(m_{DA}=\frac{7-4}{1+4}=\frac{3}{5}\); their product is also \(-1\).
5. Conclusion: All four sides are equal (\(\sqrt{34}\)), the two diagonals are equal (\(\sqrt{68}\)), and adjacent sides are perpendicular. Therefore the quadrilateral formed by the given points is a square.
6. Answer: The points \((1,7), (4,2), (-1,-1), (-4,4)\) are indeed the vertices of a square.
Question 3
Hint available
Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. Fig. 7.6
Key Idea
Three points are collinear if the slope of the line joining any two pairs of points is the same (or equivalently, the area of the triangle formed by them is zero).
Step-by-Step Solution
1. Write down the coordinates of the three points: \(A(3,1),\; B(6,4),\; C(8,6)\).
2. Compute the slope of \(AB\):
$$m_{AB}=\frac{4-1}{6-3}=\frac{3}{3}=1$$
3. Compute the slope of \(BC\):
$$m_{BC}=\frac{6-4}{8-6}=\frac{2}{2}=1$$
4. Since \(m_{AB}=m_{BC}=1\), the two slopes are equal; therefore the three points lie on the same straight line (they are collinear).
*Alternative method*: Find the area of triangle \(ABC\) using the determinant formula:
$$\Delta = \frac{1}{2}\begin{vmatrix}3&1&1\\6&4&1\\8&6&1\end{vmatrix}=0$$
The area being zero also confirms collinearity.
5. Hence, Ashima, Bharti and Camella are seated in a straight line.
2. Compute the slope of \(AB\):
$$m_{AB}=\frac{4-1}{6-3}=\frac{3}{3}=1$$
3. Compute the slope of \(BC\):
$$m_{BC}=\frac{6-4}{8-6}=\frac{2}{2}=1$$
4. Since \(m_{AB}=m_{BC}=1\), the two slopes are equal; therefore the three points lie on the same straight line (they are collinear).
*Alternative method*: Find the area of triangle \(ABC\) using the determinant formula:
$$\Delta = \frac{1}{2}\begin{vmatrix}3&1&1\\6&4&1\\8&6&1\end{vmatrix}=0$$
The area being zero also confirms collinearity.
5. Hence, Ashima, Bharti and Camella are seated in a straight line.
Question 4
Hint available
Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).
Key Idea
A point is equidistant from two given points if and only if it lies on the perpendicular bisector of the line segment joining those points. Using the distance formula, set the distances from (x, y) to each given point equal and simplify to obtain the required relation.
Step-by-Step Solution
1. Write the distance formula for the point \((x, y)\) to each of the given points:
$$\begin{aligned}
d_1 &= \sqrt{(x-7)^2+(y-1)^2},\\[4pt]
d_2 &= \sqrt{(x-3)^2+(y-5)^2}.
\end{aligned}$$
2. Equate the two distances because the point is equidistant:
$$\sqrt{(x-7)^2+(y-1)^2}=\sqrt{(x-3)^2+(y-5)^2}.$$
3. Square both sides to remove the square‑roots (both sides are non‑negative):
$$(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2.$$
4. Expand each square:
$$\begin{aligned}
(x^2-14x+49)+(y^2-2y+1) &= (x^2-6x+9)+(y^2-10y+25).
\end{aligned}$$
5. Cancel the common terms \(x^2\) and \(y^2\):
$$-14x+49-2y+1 = -6x+9-10y+25.$$
6. Collect like terms on one side:
\[-14x+6x\] + \[-2y+10y\] + \[50-34\] = 0
$$-8x+8y+16=0.$$
7. Divide by 8 to simplify:
$$-x+y+2=0.$$
8. Write the required relation between \(x\) and \(y\):
$$y = x-2.$$
$$\begin{aligned}
d_1 &= \sqrt{(x-7)^2+(y-1)^2},\\[4pt]
d_2 &= \sqrt{(x-3)^2+(y-5)^2}.
\end{aligned}$$
2. Equate the two distances because the point is equidistant:
$$\sqrt{(x-7)^2+(y-1)^2}=\sqrt{(x-3)^2+(y-5)^2}.$$
3. Square both sides to remove the square‑roots (both sides are non‑negative):
$$(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2.$$
4. Expand each square:
$$\begin{aligned}
(x^2-14x+49)+(y^2-2y+1) &= (x^2-6x+9)+(y^2-10y+25).
\end{aligned}$$
5. Cancel the common terms \(x^2\) and \(y^2\):
$$-14x+49-2y+1 = -6x+9-10y+25.$$
6. Collect like terms on one side:
\[-14x+6x\] + \[-2y+10y\] + \[50-34\] = 0
$$-8x+8y+16=0.$$
7. Divide by 8 to simplify:
$$-x+y+2=0.$$
8. Write the required relation between \(x\) and \(y\):
$$y = x-2.$$
Question 5
Hint available
Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Key Idea
Use the distance formula. For a point P(0, y) on the y‑axis, set the distances PA and PB equal and solve for y.
Step-by-Step Solution
1. Let the required point be \(P(0, y)\) because it lies on the y‑axis (\(x=0\)).
2. Write the distance from \(P\) to \(A(6,5)\):
$$PA = \sqrt{(0-6)^2 + (y-5)^2} = \sqrt{36 + (y-5)^2}.$$
3. Write the distance from \(P\) to \(B(-4,3)\):
$$PB = \sqrt{(0+4)^2 + (y-3)^2} = \sqrt{16 + (y-3)^2}.$$
4. Since \(P\) is equidistant from \(A\) and \(B\), set \(PA = PB\). Squaring both sides eliminates the square roots:
$$36 + (y-5)^2 = 16 + (y-3)^2.$$
5. Expand the squares:
$$36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9.$$
6. Cancel \(y^2\) from both sides and simplify:
$$61 - 10y = 25 - 6y.$$
7. Bring the terms containing \(y\) to one side:
$$61 - 25 = -6y + 10y \Rightarrow 36 = 4y.$$
8. Solve for \(y\):
$$y = \frac{36}{4} = 9.$$
9. Therefore the required point on the y‑axis is \(P(0, 9)\).
2. Write the distance from \(P\) to \(A(6,5)\):
$$PA = \sqrt{(0-6)^2 + (y-5)^2} = \sqrt{36 + (y-5)^2}.$$
3. Write the distance from \(P\) to \(B(-4,3)\):
$$PB = \sqrt{(0+4)^2 + (y-3)^2} = \sqrt{16 + (y-3)^2}.$$
4. Since \(P\) is equidistant from \(A\) and \(B\), set \(PA = PB\). Squaring both sides eliminates the square roots:
$$36 + (y-5)^2 = 16 + (y-3)^2.$$
5. Expand the squares:
$$36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9.$$
6. Cancel \(y^2\) from both sides and simplify:
$$61 - 10y = 25 - 6y.$$
7. Bring the terms containing \(y\) to one side:
$$61 - 25 = -6y + 10y \Rightarrow 36 = 4y.$$
8. Solve for \(y\):
$$y = \frac{36}{4} = 9.$$
9. Therefore the required point on the y‑axis is \(P(0, 9)\).
Question 6
Hint available
Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.
Key Idea
Use the section formula for internal division of a line segment.
Step-by-Step Solution
1. Let the required point be P(x, y) which divides the segment joining A(4, -3) and B(8, 5) in the ratio 3:1 internally.
2. For internal division, the section formula is
$$x = \frac{m x_2 + n x_1}{m+n}, \quad y = \frac{m y_2 + n y_1}{m+n}$$
where m:n = 3:1, (x_1, y_1) = (4, -3) and (x_2, y_2) = (8, 5).
3. Substitute the given values:
$$x = \frac{3\times8 + 1\times4}{3+1} = \frac{24+4}{4} = 7$$
$$y = \frac{3\times5 + 1\times(-3)}{3+1} = \frac{15-3}{4} = 3$$
4. Therefore, the coordinates of the required point are (7, 3).
2. For internal division, the section formula is
$$x = \frac{m x_2 + n x_1}{m+n}, \quad y = \frac{m y_2 + n y_1}{m+n}$$
where m:n = 3:1, (x_1, y_1) = (4, -3) and (x_2, y_2) = (8, 5).
3. Substitute the given values:
$$x = \frac{3\times8 + 1\times4}{3+1} = \frac{24+4}{4} = 7$$
$$y = \frac{3\times5 + 1\times(-3)}{3+1} = \frac{15-3}{4} = 3$$
4. Therefore, the coordinates of the required point are (7, 3).
Question 7
Hint available
In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?
Key Idea
Use the section formula (internal division) for coordinates of a point dividing a line segment in a given ratio. If a point P(x, y) divides AB joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n, then \(x = \frac{mx_2 + nx_1}{m+n}\) and \(y = \frac{my_2 + ny_1}{m+n}\). Solve these two equations for m and n.
Step-by-Step Solution
1. Assign symbols\
Let the required ratio be \(AP : PB = m : n\) (internal division).\
Coordinates: \(A(x_1, y_1) = (-6, 10)\), \(B(x_2, y_2) = (3, -8)\), \(P(x, y) = (-4, 6)\).\
\
2. Write the section‑formula equations\
\[\begin{aligned}
x &= \frac{m x_2 + n x_1}{m+n} = \frac{3m + (-6)n}{m+n},\\[4pt]
y &= \frac{m y_2 + n y_1}{m+n} = \frac{-8m + 10n}{m+n}.
\end{aligned}\]\
\
3. Substitute the known coordinates\
\[\begin{aligned}
-4 &= \frac{3m - 6n}{m+n} \quad\Rightarrow\quad -4(m+n) = 3m - 6n,\\[4pt]
6 &= \frac{-8m + 10n}{m+n} \quad\Rightarrow\quad 6(m+n) = -8m + 10n.
\end{aligned}\]\
\
4. Simplify each equation\
- From the first equation:\
\[-4m - 4n = 3m - 6n \Rightarrow -7m + 2n = 0 \Rightarrow 2n = 7m \Rightarrow n = \frac{7}{2}m.\
- From the second equation:\
\[6m + 6n = -8m + 10n \Rightarrow 14m - 4n = 0 \Rightarrow 4n = 14m \Rightarrow n = \frac{7}{2}m.\
Both equations give the same relation, confirming consistency.
\
5. Find the ratio\
Since \(n = \frac{7}{2}m\),\
\[\frac{m}{n} = \frac{m}{\frac{7}{2}m} = \frac{2}{7}.\]
Hence \(m:n = 2:7\).
\
6. Conclusion\
The point \((-4,6)\) divides the segment \(AB\) internally in the ratio \(AP : PB = 2 : 7\).
Let the required ratio be \(AP : PB = m : n\) (internal division).\
Coordinates: \(A(x_1, y_1) = (-6, 10)\), \(B(x_2, y_2) = (3, -8)\), \(P(x, y) = (-4, 6)\).\
\
2. Write the section‑formula equations\
\[\begin{aligned}
x &= \frac{m x_2 + n x_1}{m+n} = \frac{3m + (-6)n}{m+n},\\[4pt]
y &= \frac{m y_2 + n y_1}{m+n} = \frac{-8m + 10n}{m+n}.
\end{aligned}\]\
\
3. Substitute the known coordinates\
\[\begin{aligned}
-4 &= \frac{3m - 6n}{m+n} \quad\Rightarrow\quad -4(m+n) = 3m - 6n,\\[4pt]
6 &= \frac{-8m + 10n}{m+n} \quad\Rightarrow\quad 6(m+n) = -8m + 10n.
\end{aligned}\]\
\
4. Simplify each equation\
- From the first equation:\
\[-4m - 4n = 3m - 6n \Rightarrow -7m + 2n = 0 \Rightarrow 2n = 7m \Rightarrow n = \frac{7}{2}m.\
- From the second equation:\
\[6m + 6n = -8m + 10n \Rightarrow 14m - 4n = 0 \Rightarrow 4n = 14m \Rightarrow n = \frac{7}{2}m.\
Both equations give the same relation, confirming consistency.
\
5. Find the ratio\
Since \(n = \frac{7}{2}m\),\
\[\frac{m}{n} = \frac{m}{\frac{7}{2}m} = \frac{2}{7}.\]
Hence \(m:n = 2:7\).
\
6. Conclusion\
The point \((-4,6)\) divides the segment \(AB\) internally in the ratio \(AP : PB = 2 : 7\).
Question 8
Hint available
Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
Key Idea
Use the section formula for internal division of a line segment. If a point P divides AB in the ratio m:n (from A to P is m parts and from P to B is n parts), then \(P\bigl(\frac{n x_1 + m x_2}{m+n},\;\frac{n y_1 + m y_2}{m+n}\bigr)\). For trisection, the ratios are 1:2 and 2:1.
Step-by-Step Solution
1. Identify the end points\
\[A( x_1 , y_1 ) = (2, -2), \quad B( x_2 , y_2 ) = (-7, 4)\]
2. First point of trisection – it divides AB in the ratio \(1:2\) (one part from A, two parts from the point to B).\
Using the section formula:\
\[P_1\bigl(\frac{2\cdot x_1 + 1\cdot x_2}{1+2},\;\frac{2\cdot y_1 + 1\cdot y_2}{1+2}\bigr)\]
Substitute the coordinates:\
\[P_1\bigl(\frac{2\cdot 2 + 1\cdot (-7)}{3},\;\frac{2\cdot (-2) + 1\cdot 4}{3}\bigr)\]
\[P_1\bigl(\frac{4-7}{3},\;\frac{-4+4}{3}\bigr) = \bigl(-1,\;0\bigr)\]
3. Second point of trisection – it divides AB in the ratio \(2:1\) (two parts from A, one part from the point to B).\
Using the section formula again:\
\[P_2\bigl(\frac{1\cdot x_1 + 2\cdot x_2}{2+1},\;\frac{1\cdot y_1 + 2\cdot y_2}{2+1}\bigr)\]
Substitute the coordinates:\
\[P_2\bigl(\frac{1\cdot 2 + 2\cdot (-7)}{3},\;\frac{1\cdot (-2) + 2\cdot 4}{3}\bigr)\]
\[P_2\bigl(\frac{2-14}{3},\;\frac{-2+8}{3}\bigr) = \bigl(-4,\;2\bigr)\]
4. Result – The two points that trisect the segment AB are \((-1,0)\) and \((-4,2)\).
\[A( x_1 , y_1 ) = (2, -2), \quad B( x_2 , y_2 ) = (-7, 4)\]
2. First point of trisection – it divides AB in the ratio \(1:2\) (one part from A, two parts from the point to B).\
Using the section formula:\
\[P_1\bigl(\frac{2\cdot x_1 + 1\cdot x_2}{1+2},\;\frac{2\cdot y_1 + 1\cdot y_2}{1+2}\bigr)\]
Substitute the coordinates:\
\[P_1\bigl(\frac{2\cdot 2 + 1\cdot (-7)}{3},\;\frac{2\cdot (-2) + 1\cdot 4}{3}\bigr)\]
\[P_1\bigl(\frac{4-7}{3},\;\frac{-4+4}{3}\bigr) = \bigl(-1,\;0\bigr)\]
3. Second point of trisection – it divides AB in the ratio \(2:1\) (two parts from A, one part from the point to B).\
Using the section formula again:\
\[P_2\bigl(\frac{1\cdot x_1 + 2\cdot x_2}{2+1},\;\frac{1\cdot y_1 + 2\cdot y_2}{2+1}\bigr)\]
Substitute the coordinates:\
\[P_2\bigl(\frac{1\cdot 2 + 2\cdot (-7)}{3},\;\frac{1\cdot (-2) + 2\cdot 4}{3}\bigr)\]
\[P_2\bigl(\frac{2-14}{3},\;\frac{-2+8}{3}\bigr) = \bigl(-4,\;2\bigr)\]
4. Result – The two points that trisect the segment AB are \((-1,0)\) and \((-4,2)\).
Question 9
Hint available
Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.
Key Idea
Use the section formula (internal division) for a point dividing a line segment in a given ratio. For a point on the y‑axis its x‑coordinate is 0, which gives an equation to determine the ratio.
Step-by-Step Solution
1. Let the two given points be \(A(5,-6)\) and \(B(-1,-4)\).\
2. Suppose the y‑axis (the line \(x=0\)) cuts \(AB\) at point \(P\) which divides \(AB\) internally in the ratio \(AP:PB = m:n\).\
3. By the section formula, the coordinates of \(P\) are\
$$\left(\frac{n\cdot x_A + m\cdot x_B}{m+n},\;\frac{n\cdot y_A + m\cdot y_B}{m+n}\right).$$\
4. Since \(P\) lies on the y‑axis, its x‑coordinate is 0. Hence\
$$\frac{n\cdot 5 + m\cdot (-1)}{m+n}=0 \;\Rightarrow\; 5n - m = 0 \;\Rightarrow\; m = 5n.$$\
5. Therefore the required ratio is\
$$AP:PB = m:n = 5n:n = 5:1.$$\
6. To find the y‑coordinate of \(P\), substitute \(m=5n\) in the y‑coordinate formula:\
$$y_P = \frac{n(-6) + 5n(-4)}{5n + n}=\frac{-6n -20n}{6n}=\frac{-26n}{6n}= -\frac{13}{3}.$$\
7. Hence the point of intersection is\
$$P\;(0,\; -\frac{13}{3}).$$
2. Suppose the y‑axis (the line \(x=0\)) cuts \(AB\) at point \(P\) which divides \(AB\) internally in the ratio \(AP:PB = m:n\).\
3. By the section formula, the coordinates of \(P\) are\
$$\left(\frac{n\cdot x_A + m\cdot x_B}{m+n},\;\frac{n\cdot y_A + m\cdot y_B}{m+n}\right).$$\
4. Since \(P\) lies on the y‑axis, its x‑coordinate is 0. Hence\
$$\frac{n\cdot 5 + m\cdot (-1)}{m+n}=0 \;\Rightarrow\; 5n - m = 0 \;\Rightarrow\; m = 5n.$$\
5. Therefore the required ratio is\
$$AP:PB = m:n = 5n:n = 5:1.$$\
6. To find the y‑coordinate of \(P\), substitute \(m=5n\) in the y‑coordinate formula:\
$$y_P = \frac{n(-6) + 5n(-4)}{5n + n}=\frac{-6n -20n}{6n}=\frac{-26n}{6n}= -\frac{13}{3}.$$\
7. Hence the point of intersection is\
$$P\;(0,\; -\frac{13}{3}).$$
Question 10
Hint available
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Key Idea
In a parallelogram, the diagonals bisect each other. Hence the mid‑point of diagonal AC must be the same as the mid‑point of diagonal BD.
Step-by-Step Solution
1. Write the coordinates of the given points
\[ A(6,1),\; B(8,2),\; C(9,4),\; D(p,3). \]
2. Find the mid‑point of diagonal AC
\[ M_{AC}=\left(\frac{6+9}{2},\;\frac{1+4}{2}\right)=\left(\frac{15}{2},\;\frac{5}{2}\right)=(7.5,\;2.5). \]
3. Find the mid‑point of diagonal BD
\[ M_{BD}=\left(\frac{p+8}{2},\;\frac{3+2}{2}\right)=\left(\frac{p+8}{2},\;\frac{5}{2}\right) =\left(\frac{p+8}{2},\;2.5\right). \]
4. Equate the two mid‑points (since the diagonals bisect each other)
\[ \frac{p+8}{2}=7.5 \quad\text{and}\quad \frac{5}{2}=\frac{5}{2}. \]
The y‑coordinates are already equal; solve the x‑coordinate equation:
\[ \frac{p+8}{2}=7.5 \Rightarrow p+8=15 \Rightarrow p=7. \]
5. Conclusion
The value of \(p\) that makes \(ABCD\) a parallelogram is \(p=7\).
\[ A(6,1),\; B(8,2),\; C(9,4),\; D(p,3). \]
2. Find the mid‑point of diagonal AC
\[ M_{AC}=\left(\frac{6+9}{2},\;\frac{1+4}{2}\right)=\left(\frac{15}{2},\;\frac{5}{2}\right)=(7.5,\;2.5). \]
3. Find the mid‑point of diagonal BD
\[ M_{BD}=\left(\frac{p+8}{2},\;\frac{3+2}{2}\right)=\left(\frac{p+8}{2},\;\frac{5}{2}\right) =\left(\frac{p+8}{2},\;2.5\right). \]
4. Equate the two mid‑points (since the diagonals bisect each other)
\[ \frac{p+8}{2}=7.5 \quad\text{and}\quad \frac{5}{2}=\frac{5}{2}. \]
The y‑coordinates are already equal; solve the x‑coordinate equation:
\[ \frac{p+8}{2}=7.5 \Rightarrow p+8=15 \Rightarrow p=7. \]
5. Conclusion
The value of \(p\) that makes \(ABCD\) a parallelogram is \(p=7\).