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Coordinate Geometry
CH07 Question Bank
CBSE_CH07_QUESTION_BANK
Grade 10
Question:
[Case Study]
Two villages, A and B, are marked on a district map using a coordinate grid (1 unit = 1 km), with A at $(2,3)$ and B at $(14,19)$. The district authority wants to build a rest stop $P$ on the straight road joining A and B, such that $P$ divides $AB$ in the ratio $1:3$ (measuring from A).
(a) Find the straight-line distance between villages A and B. [1 Mark]
(b) Find the coordinates of the rest stop P. [1 Mark]
(c) Find the distance AP. [1 Mark]
(d) Verify that AP is exactly one-quarter of AB, consistent with the given ratio 1:3. [1 Mark]
Two villages, A and B, are marked on a district map using a coordinate grid (1 unit = 1 km), with A at $(2,3)$ and B at $(14,19)$. The district authority wants to build a rest stop $P$ on the straight road joining A and B, such that $P$ divides $AB$ in the ratio $1:3$ (measuring from A).
(a) Find the straight-line distance between villages A and B. [1 Mark]
(b) Find the coordinates of the rest stop P. [1 Mark]
(c) Find the distance AP. [1 Mark]
(d) Verify that AP is exactly one-quarter of AB, consistent with the given ratio 1:3. [1 Mark]
Step-by-Step Solution
Key Concept: Case study on coordinate geometry.
(a) Find the straight-line distance between villages A and B. [1 Mark]
$AB=\sqrt{(14-2)^2+(19-3)^2}=\sqrt{144+256}=\sqrt{400}=20$ km. [1.0 Mark]
(b) Find the coordinates of the rest stop P. [1 Mark]
$P_x=\dfrac{1(14)+3(2)}{1+3}=\dfrac{14+6}{4}=5$; $P_y=\dfrac{1(19)+3(3)}{4}=\dfrac{19+9}{4}=7$. So $P=(5,7)$. [1.0 Mark]
(c) Find the distance AP. [1 Mark]
$AP=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5$ km. [1.0 Mark]
(d) Verify that AP is exactly one-quarter of AB, consistent with the given ratio 1:3. [1 Mark]
$\dfrac{AP}{AB}=\dfrac{5}{20}=\dfrac14$, consistent with $AP:PB=1:3$ (i.e. $AP$ is $1$ part out of the total $4$ parts). Verified. [1.0 Mark]
$AB=\sqrt{(14-2)^2+(19-3)^2}=\sqrt{144+256}=\sqrt{400}=20$ km. [1.0 Mark]
(b) Find the coordinates of the rest stop P. [1 Mark]
$P_x=\dfrac{1(14)+3(2)}{1+3}=\dfrac{14+6}{4}=5$; $P_y=\dfrac{1(19)+3(3)}{4}=\dfrac{19+9}{4}=7$. So $P=(5,7)$. [1.0 Mark]
(c) Find the distance AP. [1 Mark]
$AP=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5$ km. [1.0 Mark]
(d) Verify that AP is exactly one-quarter of AB, consistent with the given ratio 1:3. [1 Mark]
$\dfrac{AP}{AB}=\dfrac{5}{20}=\dfrac14$, consistent with $AP:PB=1:3$ (i.e. $AP$ is $1$ part out of the total $4$ parts). Verified. [1.0 Mark]
Correct Answer: $AB=\sqrt{(14-2)^2+(19-3)^2}=\sqrt{144+256}=\sqrt{400}=20$ km. [1.0 Mark] | $P_x=\dfrac{1(14)+3(2)}{1+3}=\dfrac{14+6}{4}=5$; $P_y=\dfrac{1(19)+3(3)}{4}=\dfrac{19+9}{4}=7$. So $P=(5,7)$. [1.0 Mark] | $AP=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5$ km. [1.0 Mark] | $\dfrac{AP}{AB}=\dfrac{5}{20}=\dfrac14$, consistent with $AP:PB=1:3$ (i.e. $AP$ is $1$ part out of the total $4$ parts). Verified. [1.0 Mark]
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