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Straight Lines and Pair of Straight Lines question

2020 · 4 Sep · Shift 1 · Q28

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Straight Lines and Pair of Straight Lines
2020 · 4 Sep · Shift 1 · Q28

Straight Lines and Pair of Straight Lines question

2020 · 4 Sep · Shift 1 · Q28

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1

Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is :

A
10/3
B
5
C
20/3
D
6

View written solutionFree

Correct answer: D

Set up coordinates

Let the horizontal ground line $AC$ be the $x$ -axis. Choose:

$A=(0,0)$
$B=(0,15)$ since $AB=15$
$C=(d,0)$
$D=(d,10)$ since $CD=10$

Here $d$ is the horizontal distance between the poles.

We need the height of the intersection point $P$ of lines $BC$ and $AD$ above line $AC$ .

Equation of line $AD$

Line $AD$ passes through $(0,0)$ and $(d,10)$ . So its slope is $m_{AD}=\frac{10-0}{d-0}=\frac{10}{d}.$

Hence its equation is $y=\frac{10}{d}x.$

Equation of line $BC$

Line $BC$ passes through $(0,15)$ and $(d,0)$ . So its slope is $m_{BC}=\frac{0-15}{d-0}=-\frac{15}{d}.$

Hence its equation is $y=15-\frac{15}{d}x.$

Find the intersection point $P$

At intersection, the $y$ -values are equal: $\frac{10}{d}x=15-\frac{15}{d}x.$

So, $\frac{25}{d}x=15$ $x=\frac{15d}{25}=\frac{3d}{5}.$

Now substitute into $y=\frac{10}{d}x$ : $y=\frac{10}{d}\cdot \frac{3d}{5}=6.$

Thus the intersection point $P$ is at height $6\text{ m}$ above the line $AC$ .

Check options

A: $\frac{10}{3}$ ✗
B: $5$ ✗
C: $\frac{20}{3}$ ✗
D: $6$ ✓

So the correct option is D.

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