Question

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Answer 1

Answer:

Step-by-step explanation:

Segments AB and BC intersect each other at 90° at B.

Let the equation of the segment AB → y = mx + b

Here, m = Slope of the line

b = y-intercept

Slope of the line AB passing through A(14, -1) and B(2, 1)

Slope = $\frac{y_2-y_1}{x_2-x_1}$

          = $\frac{1+1}{2-14}$

          = $-\frac{2}{12}$

          = $-\frac{1}{6}$

Equation of the line will be,

$y=-\frac{1}{6}(x)+b$

Since, AB passes through (2, 1)

$1=-\frac{1}{6}(2)+b$

$b=1+\frac{1}{3}$

$b=\frac{4}{3}$

Therefore, y-intercept of AB = $\frac{4}{3}$

Equation of AB → $y=-\frac{1}{6}(x)+\frac{4}{3}$

Since, line BC is perpendicular to AB,

By the property of perpendicular lines,

$m_1\times m_2=-1$

Here, $m_1$ and $m_2$ are the slopes of line AB and BC respectively.

By this property,

$-\frac{1}{6}\times m_2=-1$

$m_2=6$

Equation of a line passing through a point (h, k) and slope 'm' is,

(y - k) = m(x - h)

Therefore, equation of line BC passing through B(2, 1) and slope = 6,

y - 1 = 6(x - 2)

y = 6x - 11

Since, line BC passes through C(x, 13),

13 = 6x - 11

6x = 24

x = 4

Therefore, x-coordinate of point C will be, x = 4