Question

What is the equation in point-slope form of a line that passes through the points (5, −3)(5, −3) and (−2, 9)(−2, 9) ? y−3=−2(x 5)y−3=−2(x 5) y 3=−127(x−5)y 3=−127(x−5) y 3=−2(x−5)y 3=−2(x−5) y−3=−127(x 5)

Answer 1

(5,-3)(-2,9)
slope(m) = (y2 - y1) / (x2 - x1)
slope(m) = (9 - (-3) / (-2 - 5) = (9 + 3) / -7 = - 12/7

y - y1 = m(x - x1)....point slope form
slope(m) = -12/7
(5,-3)...x1 = 5 and y1 = -3
now we sub...and pay close attention to ur signs
y - (-3) = -12/7(x - 5).....we are not done yet...
y + 3 = -12/7(x - 5) <== ur answer

Answer 2

The equation in point-slope form of a line that passes through the points (5, -3) and (-2, 9) is $y+3= \frac{-12}{7} (x - 5)$

For a line that passes through the points (x₁, y₁) and (x₂, y₂), the point-slope form of the equation of the line is:

$y - y_1=m(x-x_1)$

where m is the slope given by the formula:

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

The line passes through the points (5, -3) and (-2, 9)

That is, x₁ = 5, y₁ = -3, x₂ = -2, y₂ =  9

Calculate the slope (m) by substituting x₁ = 5, y₁ = -3, x₂ = -2, y₂ =  9

into the equation $m = \frac{y_2-y_1}{x_2-x_1}$

$m = \frac{9-(-3)}{-2-5} \\\\m = \frac{-12}{7}$

Substitute x₁ = 5, y₁ = -3, and m = -12/7 into the line equation

$y - y_1=m(x-x_1)$

$y - (-3) = \frac{-12}{7} (x - 5)\\\\y+3= \frac{-12}{7} (x - 5)$

Therefore, the equation in point-slope form of a line that passes through the points (5, -3) and (-2, 9) is $y+3= \frac{-12}{7} (x - 5)$

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