BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1, Problem 126RE
To determine

To calculate:The equation for the line passing through the point (1,7) and also seen to be perpendicular to a line with equation x3y+16=0 .

Expert Solution

Answer to Problem 126RE

The equation for the line passing through the point (1,7) and perpendicular to a line x3y+16=0 is y=3x+10 .

Explanation of Solution

Given information:

A line passing through the point P1=(1,7) and perpendicular to a line x3y+16=0

Formula used:

Slope m of the line passing through two points in general say P1=(x1,y1) and P2=(x2,y2) is:

  m=y2y1x2x1

Point-slope equation for a given line passing along point P1=(x1,y1) is:

  yy1=m(xx1)

Slope-intercept equation for a given line which has slope as m and y −intercept as b is:

  y=mx+b

Two-intercept equation for a given line which has x -intercept as a and y −intercept as b is:

  xa+yb=1

When two lines are perpendicular then the product of their slopes is zero that is m1.m2=1

Calculation:

Consider thegiven point P1=(1,7) and line x3y+16=0

Recall the slope-intercept equation for a given line which has slope as m and y −intercept as b is:

  y=mx+b

Let the line passing through point P1=(1,7) and perpendicular to line x3y+16=0 has the general form:

  y=m1x+b1

Where m1 is the slope and b1 is the y -intercept of the line.

Now, we try to express the line x3y+16=0 in the same form:

Firstly let its general form be y=m2x+b2

Where m2 is the slope and b2 is the y -intercept of the line.

  x3y+16=03y=x16y=x163y=(13)x+163

Comparing with y=m2x+b2 to get:

  m2=13b2=163

As it’s given that line y=m1x+b1 and line y=m2x+b2 are perpendicular.

Therefore, recall that when two lines are perpendicular then the product of their slopes is zero that is m1.m2=1

That is:

  m1.13=1m1=3

Therefore, equation of the required line reduces to:

  y=3x+b1(1)

Also it is given that this line passes through P1=(1,7) .

Then in order to proceed following should be known:

  x=1y=7

Putting this in equation (1) to get:

  7=3(1)+b1b1=7+3b1=10

Put this value of b1 in (1) to get the required equation as:

  y=3x+10

Hence, the equation for the line passing through the point (1,7) and perpendicular to a line x3y+16=0 is y=3x+10 .

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