MAH-CET 2026 — Mathematics PYQ
MAH-CET | Mathematics | 2026The equations to straight lines passing through the points and equally inclined to the lines and are
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The equations to straight lines passing through the points (4,5) and equally inclined to the lines 3x=4y+7 and 5y=12x+6 are
9x−7y=1 and 7x+9y=73
7x−9y=1 and 7x+9y=73
9x−7y=1 and 9x+7y=73
7x−7y=1 and 9x+9y=73
9x−7y=1 and 7x+9y=73
Step 1: Understand the geometric condition
Any line that is equally inclined to two intersecting lines must be parallel to one of the angle bisectors of those two lines. Therefore, the slopes of our required lines will match the slopes of the angle bisectors of the given reference lines.
Let us rewrite the given lines in standard form Ax+By+C=0:
Line 1 (u1): 3x−4y−7=0
Line 2 (u2): 12x−5y+6=0
Step 2: Find the equations of the angle bisectors
The formula for the equations of the angle bisectors between two lines A1x+B1y+C1=0 and A2x+B2y+C2=0 is:
A12+B12A1x+B1y+C1=±A22+B22A2x+B2y+C2
Substituting our values:
32+(−4)23x−4y−7=±122+(−5)212x−5y+6
253x−4y−7=±16912x−5y+6
53x−4y−7=±1312x−5y+6
Let's compute both cases to find the slopes of the bisectors:
Case 1: Taking the positive sign (+)
13(3x−4y−7)=5(12x−5y+6)
39x−52y−91=60x−25y+30
Rearranging all terms to one side:
21x+27y+121=0
The slope of this bisector line (m1) is:
m1=−Coefficient of yCoefficient of x=−2721=−97
Case 2: Taking the negative sign (-)
13(3x−4y−7)=−5(12x−5y+6)
39x−52y−91=−60x+25y−30
Rearranging all terms to one side:
99x−77y−61=0
Dividing the coefficients by 11:
9x−7y−1161=0
The slope of this bisector line (m2) is:
m2=−Coefficient of yCoefficient of x=−−79=79
Step 3: Formulate equations passing through (4,5)
Now, we find the equations of the required lines using the point-slope form y−y1=m(x−x1) with the point (4,5).
Line with slope m1=−97:
y−5=−97(x−4)
9(y−5)=−7(x−4)
9y−45=−7x+28
7x+9y=73
Line with slope m2=79:
y−5=79(x−4)
7(y−5)=9(x−4)
7y−35=9x−36
9x−7y=1
Combining our results, the two lines are 9x−7y=1 and 7x+9y=73.
(a) 9x−7y=1 and 7x+9y=73
Step 1: Understand the geometric condition
Any line that is equally inclined to two intersecting lines must be parallel to one of the angle bisectors of those two lines. Therefore, the slopes of our required lines will match the slopes of the angle bisectors of the given reference lines.
Let us rewrite the given lines in standard form Ax+By+C=0:
Line 1 (u1): 3x−4y−7=0
Line 2 (u2): 12x−5y+6=0
Step 2: Find the equations of the angle bisectors
The formula for the equations of the angle bisectors between two lines A1x+B1y+C1=0 and A2x+B2y+C2=0 is:
A12+B12A1x+B1y+C1=±A22+B22A2x+B2y+C2
Substituting our values:
32+(−4)23x−4y−7=±122+(−5)212x−5y+6
253x−4y−7=±16912x−5y+6
53x−4y−7=±1312x−5y+6
Let's compute both cases to find the slopes of the bisectors:
Case 1: Taking the positive sign (+)
13(3x−4y−7)=5(12x−5y+6)
39x−52y−91=60x−25y+30
Rearranging all terms to one side:
21x+27y+121=0
The slope of this bisector line (m1) is:
m1=−Coefficient of yCoefficient of x=−2721=−97
Case 2: Taking the negative sign (-)
13(3x−4y−7)=−5(12x−5y+6)
39x−52y−91=−60x+25y−30
Rearranging all terms to one side:
99x−77y−61=0
Dividing the coefficients by 11:
9x−7y−1161=0
The slope of this bisector line (m2) is:
m2=−Coefficient of yCoefficient of x=−−79=79
Step 3: Formulate equations passing through (4,5)
Now, we find the equations of the required lines using the point-slope form y−y1=m(x−x1) with the point (4,5).
Line with slope m1=−97:
y−5=−97(x−4)
9(y−5)=−7(x−4)
9y−45=−7x+28
7x+9y=73
Line with slope m2=79:
y−5=79(x−4)
7(y−5)=9(x−4)
7y−35=9x−36
9x−7y=1
Combining our results, the two lines are 9x−7y=1 and 7x+9y=73.
(a) 9x−7y=1 and 7x+9y=73