Question

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

• A. $-5$
• B. $-\frac{35}{3}$
• C. $\frac{35}{3}$
• D. $5$

Answer 1

The correct option is A $-5$

Given $2x-3y+17=0$ ……………. $\left(1\right)$

is a line perpendicular to line passing through point $P(7,17)$ & $Q(15,\beta )$ then

equation of line passing throught P & Q is

$y-17=\left(\frac{\beta -17}{15-7}\right)(x-7)$

$8y-136=\beta x-7\beta -17x+119$

$(7-\beta )x+8y-255=0$ ………... $\left(2\right)$

Line $\left(1\right)$ & $\left(2\right)$ are perpendicular to each other then

$m_1{m}_2=1$

Product of slopes of line $\left(1\right)$ & $\left(2\right)$ 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 $[\beta =-5]$.