Question

The coordinates of A, B, and C in the diagram are A(p,4), B(6,1), and C(9,q). Which equation correctly relates p and q? Hint: Since is perpendicular to , the slope of × the slope of = -1. A. q − p = 7 B. p + q = 7 C. p − q = 7 D. -q − p = 7

Answer 1

Answer:

slope of AB = -3/ 6-p

and, slope of BC = q-1/3

Now for perpendicular lines

$\frac{ - 3}{6 - p} \times \frac{q - 1}{3} = - 1$

or, -(q-1) = -(6-p)

or, q-1 = 6-p

or, p+q =7

option B is correct...

Answer 2

According to given details in question

We have to find the slope of given coordinates.

Step-by-step explanation:

Given,

A,B,C are points with their coordinates .

A(p,4), B(6,1), and C(9,q)

Step-1:

The formula of slope

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

Step-2:

• Slope of AB in which A(p,4), B(6,1)

Here, x1= p, y1= 4, x2= 6, y2= 1

$Slope \: of \: AB = \frac{1 - 4}{6 - p}$

Subtract the terms ,

$Slope \: of \: AB \: or \: m_1= \frac{ - 3}{6 - p} \: \: \: \: ...(1)$

• Slope of BC in which B(6,1), and C(9,q).

Here, x1= 6, y1= 4, x2= 9, y2= q

$Slope \: of \: BC = \frac{q-1}{9-6}$

Subtract the terms ,we get

$Slope \: of \: BC \: or \: m_2= \frac{q-1}{3} \: \: \: \: ....(2)$

Step-3:

From Hint, the slopes are perpendicular lines

Formula:

$m_1×m_2 = -1$

Put the values from (1) and (2)

$\frac{ - 3}{6 - p} \times \frac{q - 1}{3} = - 1$

Multiply the terms and divide cross ,we get

$\frac{ - 1}{6 - p} \times \frac{q - 1}{1} = - 1$

$\frac{ - (q - 1)}{6 - p} = - 1$

Shift the denominator toward right hand side,

$- ({q - 1}) = - 1 \times( 6 - p)$

Solve the terms

$q + 1 = - 6 + p$

or

$p + q = 1 + 6$

$p + q = 7$

Hence ,

The slope in terms of p and q is p+q=7.

Option (B) p + q = 7 is correct.

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