Question

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

• A. -5
• B. 29
• C. 5
• D. -29

Answer 1

The Correct Option is C

To determine the value of β, we first need to find the slope of the line passing through the points (7, 17) and (15, β).
The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by:
\(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\)

For the given points (7, 17) and (15, β), the slope is:
\(m = \frac{{\beta - 17}}{{15 - 7}} = \frac{{\beta - 17}}{8}\)

Now, we know that a line is perpendicular to another line if and only if the product of their slopes is -1. 
Therefore, we need to find the slope of the line \(2x - 3y + 17 = 0\) and determine the value of β that makes the product of the slopes -1.

The given line \(2x - 3y + 17 = 0\) can be rewritten in slope-intercept form as:
\(y = \frac{2}{3}x + \frac{17}{3}\)
Comparing this equation with the standard slope-intercept form \(y = mx + b\), we can see that the slope of this line is \(\frac{2}{3}\)
Now, we have:
slope of the line passing through (7, 17) and \((15, \beta) = \frac{{\beta - 17}}{8}\)
slope of the line \(2x - 3y + 17 = 0\)

To find β, we set the product of the slopes equal to -1:
\(\frac{{\beta - 17}}{8} \times \frac{2}{3} = -1\)
Simplifying the equation:
\((\beta - 17) \times \frac{2}{3}\)
Multiplying both sides by \(\beta = \frac{3}{2} + 17\)
\((\beta - 17) = -8 \times \frac{3}{2}\)(β - 17) = -8 * (3/2) 
\(= -12\)
Solving for β:
\(β = -12 + 17 = 5\)
Therefore, β equals 5 (option C).