CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

The straight lines 3x+4y=5 and 4x3y=15 intersect at the point A. On these lines points B and C are chosen so that AB=AC.Possible equation(s) of the line BC passing through (1,2) is/are

A
x+7y=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y+7x=9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
7xy=13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7yx=13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
B y+7x=9
D 7yx=13
The lines 3x+4y=5 and 4x3y=15 respectively.
Hence the two given straight lines are at right angles.

So, according to the condition AB=AC there are 2 such lines are possible which make an angle of 45° with both lines
Now slope of the line which makes an angle of 45° with 4x3y+5=0is
m=4/3±114/3=7,17
Hence the required equations are
y2=7(x1) and y2=17(x1)
y+7x=9 and 7yx=13

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tango With Straight Lines !!
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon