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Q. If the straight line $2 x-3 y+17=0$ is perpendicular to the line passing through the points $(7,17)$ and $(15, \beta)$, then $\beta$ equals

KCETKCET 2022Straight Lines

Solution:

Given $2 x-3 y+17=0$
is a line perpendicular to line passing through point $P (7,17) \& Q (15, \beta)$ then equation of line passing throught $P \& Q$ is
$y-17=\left(\frac{\beta-17}{15-7}\right)(x-7)$
$8 y-136=\beta x-7 \beta-17 x+119$
$(7-\beta) x+8 y-255=0 \ldots \ldots \ldots . .(2)$
Line (1) \& (2) are perpendicular to each other then
$m _{1} m _{2}=1$
Product of slopes of line (1) \& (2) is $-1$
Now,
$m _{1}=\frac{2}{3}$ and $m _{2}=\frac{\beta-7}{8}$
$m _{1} m _{2}=-1 \Rightarrow \frac{2}{3}\left(\frac{\beta-7}{48}\right)=-1$
$ \beta-7=-12 $
$ \beta=-12+7=-5 $
Hence $[\beta=-5]$.