Q. 55.0( 3 Votes )
Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.
Answer :
Given:
Lines 2x + 3y = 21 and 3x – 4y + 11 = 0
To find:
The distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.
Concept Used:
Distance of a point from a line.
Explanation:
Solving the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 we get:
⇒ x = 3, y = 5
So, the point of intersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).
Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is
d
Hence, distance is .
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PREVIOUSShow that the perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.NEXTFind the length of the perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the lines 2x – 3y + 14 = 0 and 5x + 4y – 7 = 0.
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