# The lines 3 x − 5 y + 19 = 0 and 10 x + 6 y − 50 = 0 are perpendicular and also find their point of intersection.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter B, Problem 42E
To determine

## To show: The lines 3x−5y+19=0 and 10x+6y−50=0 are perpendicular and also find their point of intersection.

Expert Solution

The point of the intersection is (2,5).

### Explanation of Solution

Formula used:

The two lines with slopes m1 and m2 are perpendicular if and only if m1m2=1.

Calculation:

It is enough to show that m1m2=1, in order to say the two lines are perpendicular.

Find the slope of the line 3x5y+19=0 as follows.

3x5y+19=05y=193xy=19535x

y=35x195. (1)

Therefore, the slope is m1=35.

Find the slope of the line 10x+6y50=0 as follows.

10x+6y50=06y=5010xy=506106x

y=53x+253. (2)

Therefore, the slope is m2=53.

Here,

m1m2=35×53m1m2=1.

From the given formula, the given two lines are perpendicular.

Hence, it is proved that the lines 3x5y+19=0 and 10x+6y50=0 are perpendicular.

In order to find the intersection of their points, equate the equations (1) and (2) as follows.

35x+195=53x+25335x+53x=195+2539x+25x15=57+125159x+25x=57+125

On further simplification,

34x=68x=6834x=2

Now substitute x=2 in either 3x5y+19=0 or 10x+6y50=0.

3(2)5y+19=065y=195y=196y=255y=5

Therefore, the point of the intersection is (2,5).

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