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McGraw Hill Glencoe Geometry, 2012

MH

McGraw Hill Glencoe Geometry, 2012 View details

1. Bisectors of Triangles

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Exercise 65 Page 333

Find a perpendicular line through the given point. Then, find the intersection point between both lines. Finally, find the distance between the intersection point and the given one.

$5$

Practice makes perfect

To find the distance between $y = 2 x + 2$ and $(- 1, - 5),$ we begin by finding a perpendicular line from $(- 1, - 5) .$

Finding the Perpendicular Line

The slope of the given line is

2

and, therefore, the slope of the perpendicular line is

- \frac{1}{2} .

Given that the perpendicular line passes through

(- 1, - 5),

we can find its equation by starting with the point-slope form.

y - y_{1} = m (x - x_{1})

SubstituteValues

Substitute values

y - (- 5) = - \frac{1}{2} (x - (- 1))

▼

Write in slope-intercept form

$a - (- b) = a + b$

y + 5 = - \frac{1}{2} (x + 1)

Distribute $- \frac{1}{2}$

y + 5 = - \frac{1}{2} x - \frac{1}{2}

$LHS - 5 = RHS - 5$

y = - \frac{1}{2} x - \frac{1}{2} - 5

$a = \frac{2 \cdot a}{2}$

y = - \frac{1}{2} x - \frac{1}{2} - \frac{1 0}{2}

Subtract fractions

y = - \frac{1}{2} x - \frac{1 1}{2}

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Subchapter links

Check Your Understanding p.329

Practice and Problem Solving p.329-332

H.O.T. Problems p.332

Standardized Test Practice p.333

Spiral Review p.333

Skills Review p.333