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danni2528
Find the slope‐intercept equation of the line with the following properties: Perpendicular to the line x- 4y = 2; containing the point (5,2).
you have any ideas how to start ?
@danni2528 Do you know how to find slope of a line from its equation?
Slope-Intercept form is: y = mx + b m is slope b is y-intercept ------------------ Change into Slope-Intercept form. x - 4y = 2 Subtract x from both sides. -4y = -x + 2 Divide both sides by -4. y = 1/4x - 0.5 So you know the slope of this line is 1/4. *****The slope of Perpendicular lines is the opposite reciprocal. So the new line will have a slope of -4/1 or -4. ----------------------- So far, we have the slope of the new line (-4) and we have a given point (5, 2). Use Point-Slope formula. \[y - y1 = m(x - x1) \]\[y - 2 = -4(x - 5) \]Change into Slope-Intercept form. \[y - 2 = -4(x - 5) \]Distribute.\[y - 2 = -4x + 20\]Add 2 to both sides.\[y = -4x + 22\]^There is your equation for a line perpendicular to x- 4y = 2.