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ab and bc form a right angle at their point of intersection, b.
if the coordinates of a and bare (14, -1) and (2, 1), respectively, the y-intercept of ab is
and the equation of bc
is y =
x+
if the y-coordinate of point c is 13, its x-coordinate is

y-intercept of AB is 4/3.

The equation of line BC is y=6x-11.

The x-coordinate of point C is 4.

Step-by-step explanation:

Given information: AB⊥BC, A(14,-1) and B(2,1).

If a line passes through two points then the equation of line is

where, m is slope.

Slope of line AB is

The equation of line AB is

The equation of line AB is .

Substitute x=0 to find the y-intercept of AB.

Therefore y-intercept of AB is 4/3.

Line AB and BC are perpendicular, so product of their slopes is -1.

Slope of BC is 6 and it passes through (2,1). So equation of BC is

Therefore the equation of line BC is y=6x-11.

Point C lies on line BC. The y-coordinate of C is 13.

Substitute y=13 in the above equation.

Add 11 both sides.

Divide both sides by 6.

Therefore the x-coordinate of point C is 4.

answdf

step-by-step explanation: bcuz d is right

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step-by-step explanation: