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

Answer: (C)

5

Answer 2

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).