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Pair of Linear Equations in Two Variables
CH03 Question Bank
CBSE_CH03_QUESTION_BANK
Grade 10
Question:
Solve the pair of equations $x-y+1=0$ and $3x+2y-12=0$ graphically. Also find the coordinates of the vertices of the triangle formed by these two lines and the $x$-axis, and hence find the area of this triangle.
Step-by-Step Solution
Key Concept: Plot both lines using a table of values, read off the intersection point as the solution, then find where each line meets the $x$-axis to identify the triangle's vertices.
For $x-y=-1$: when $x=-1,y=0$; when $x=2,y=3$. For $3x+2y=12$: when $x=4,y=0$; when $x=2,y=3$. Plotting both lines on the same graph. [1.0 Mark]
The two lines intersect at $(2,3)$, so the solution is $x=2,\ y=3$. [1.0 Mark]
Line $x-y=-1$ meets the $x$-axis at $(-1,0)$ (setting $y=0$); line $3x+2y=12$ meets the $x$-axis at $(4,0)$ (setting $y=0$). [1.0 Mark]
So the triangle has vertices $(2,3),\ (-1,0),\ (4,0)$. Its base lies along the $x$-axis, from $(-1,0)$ to $(4,0)$, so base $=4-(-1)=5$ units; the height is the $y$-coordinate of the third vertex $=3$ units. [1.0 Mark]
Area $=\dfrac12\times\text{base}\times\text{height}=\dfrac12\times5\times3=7.5$ square units. [1.0 Mark]
The two lines intersect at $(2,3)$, so the solution is $x=2,\ y=3$. [1.0 Mark]
Line $x-y=-1$ meets the $x$-axis at $(-1,0)$ (setting $y=0$); line $3x+2y=12$ meets the $x$-axis at $(4,0)$ (setting $y=0$). [1.0 Mark]
So the triangle has vertices $(2,3),\ (-1,0),\ (4,0)$. Its base lies along the $x$-axis, from $(-1,0)$ to $(4,0)$, so base $=4-(-1)=5$ units; the height is the $y$-coordinate of the third vertex $=3$ units. [1.0 Mark]
Area $=\dfrac12\times\text{base}\times\text{height}=\dfrac12\times5\times3=7.5$ square units. [1.0 Mark]
Correct Answer:
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