Question

The straight line 2x-3y+17=0 is perpendicular to the line passing through the points
\[\left( 7,\text{ }17 \right)\text{ and (15, }\beta \text{) then find }\beta \text{?}\]
(a)-5
\[(b)-\dfrac{35}{3}\]
\[(c)\dfrac{35}{3}\]
(d) 5

Answer 1

Hint: First find the slopes of lines using the 2 formulas given below and then name the slopes as a and b. Use the conditions if two straight lines with slope a, b are perpendicular then
\[a\times b=-1\].
Use this for the first line. If the slope of a line with equation: ax + by + c = 0 is m, then:
\[m=-\dfrac{a}{b}\]
Use this for the second line. If the slope of a line passing through the points (x, y) and (a, b) is m, then:
\[m=\dfrac{b-y}{a-x}\]

Complete step-by-step answer:
So firstly, we need to find slopes of both the lines.
Let us assume:
\[\begin{align}
  & AB\equiv 2x-3y+17=0 \\
 & CD\equiv \text{line passing through the point }\left( 7,17 \right)\text{ and }\left( 15,\beta \right) \\
\end{align}\]
First we will find the slope of line AB.
Use the condition:
If the slope of a line with equation: ax + by + c = 0 is m, then:
\[m=-\dfrac{a}{b}\]
Let the slope of AB be p.
So applying above condition to AB ( 2x - 3y + 17 = 0),
From AB we can say a=2 and b=-3.
Substituting values of a, b in the slope condition, we get:
\[p=-\dfrac{2}{-3}=\dfrac{2}{3}.....\left( 1 \right)\]
If the slope of a line passing through the points (x, y) and (a, b) is m, then:
\[m=\dfrac{b-y}{a-x}.....\left( 2 \right)\]
By using the above condition we can find the slope of CD.
Let the slope be q.
By looking at CD we can say that:
x = 7,
y = 17,
a = 15,
\[b=\beta \],
By substituting values of x, y, a, b, q into the condition (2), we get:
\[q=\dfrac{b-y}{a-x}=\dfrac{\beta -17}{15-7}=\dfrac{\beta -17}{8}.....\left( 3 \right)\]
We know by condition of perpendicular lines:
If two straight lines with slope a, b are perpendicular then
\[a\times b=-1\]
By applying above condition, we get:
\[p\times q=-1.....\left( 4 \right)\]
By substituting equation (3) equation (2) into equation (4), we get:
\[\dfrac{2}{3}\times \dfrac{\beta -17}{8}=-1\]
By cross multiplying, we get:
\[\begin{align}
  & 2\left( \beta -17 \right)=-24 \\
 & 2\beta -34=-24 \\
\end{align}\]
By simplifying, we get:
\[\begin{align}
  & 2\beta =10 \\
 & \beta = 5 \\
\end{align}\]
\[\therefore \]Option (d) is correct.

Note: Be careful with negative signs, as the slope formula of the first line already has a negative sign if you confuse it with the wrong answer.