CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If the point (2k-3,k+2) lies on the line 2x+3y+15=0, find the value of k.


A

k=-157

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

k=-137

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

k=157

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

k=137

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

k=-157


Explanation for the correct option:

Point will lie on line

The point (2k-3,k+2) lies on the line 2x+3y+15=0,

This means that the point will satisfy the equation.

So, if we put the value of x=2k-3 and y=k+2 in the equation 2x+3y+15=0, then, we must get the RHS as zero.

LHS=2(2k-3)+3(k+2)+15=4k-6+3k+6+15=7k+15

Equating LHSandRHS

7k+15=0.7k=-15k=-157

Hence, the value of k=-157.

Hence, option (A) is the correct answer


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Theorems
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon