Question

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

• A. $-5$
• B. $29$
• C. $5$
• D. $-29$

Answer 1

Answer: (A)

Given $2x-3y+17=0$
is a line perpendicular to line passing through point $P(7,17)\&Q(15,\beta )$ then equation of line passing throught $P\&Q$ is
$y-17=\left(\frac{\beta -17}{15-7}\right)(x-7)$
$8y-136=\beta x-7\beta -17x+119$
$(7-\beta )x+8y-255=0\dots \dots \dots ..(2)$
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 $[\beta =-5]$.