The point Q has coordinates (2,3)
The point R has coordinates (a, b)
A line perpendicular to QR is given by the equation 5x + 2y = 9
Find an expression for b in terms of a. W

Answers

Answer 1

Answer:

b= (2a+11)/5

Step-by-step explanation:

5x + 2y = 9

2y=9-5x

y= 9/2 - 5x/2

slope, m1= -5/2

for perpendicularity, m1 × m2 = -1

slope for QR= m2=

[tex] = - 1 \div \frac{ - 5}{2} = - 1 \times \frac{2}{ - 5} = \frac{ - 2}{ - 5} = \frac{2}{5} [/tex]

m2 is 2/5

[tex]qr = \frac{b - 3}{a - 2} [/tex]

[tex] \frac{2}{5} = \frac{b - 3}{a - 2} [/tex]

5(b-3)=2(a-2)

5b-15=2a-4

5b=2a-4+15

5b= 2a+11

[tex]b = \frac{2a + 11}{5} [/tex]

Answer 2

An expression for b in terms of a is b = (2a + 11)/5.

What are lines and their slopes?

We know lines have various types of equations, the general type is

Ax + By + c = 0, and the equation of a line in slope-intercept form is

y = mx + b.

Where slope = m and b = y-intercept.

the slope is the rate of change of the y-axis with respect to the x-axis and the y-intercept is the (0,b) where the line intersects the y-axis at x = 0.

Given an equation of a line 5x + 2y = 9 ⇒ 2y = - 5x + 9 ⇒ y = (-5/2)x + 9/2.

So, the slope of the line parallel to QR is the slope negative reciprocal of (-5/2) which is (2/5).

We know slope m = (b - 3)/(a - 2).

∴ (2/5) = (b - 3).(a - 2).

2(a - 2) = 5(b - 3).

2a - 4 = 5b - 15.

2a + 11 = 5b.

b = (2a + 11)/5.

learn more about lines and slopes here :

https://brainly.com/question/3605446

#SPJ6


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