2. Find the equation of a straight line passing through the points A(7,5) and B(-9,5)
1. Find the equation of a straight line passing through the points A(7,5) and B(-9,5)
Solution:
The given points are A(7,5),B(-9,5)
:. x_1=7,y_1=5,x_2=-9,y_2=5
Using two-points formula, The equation of a line AB is
(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)
:. (y-5)/(5-5)=(x-7)/(-9-7)
:. (y-5)/(0)=(x-7)/(-16)
:. (y-5)/(0)=(x-7)/(-1)
:. -1(y-5)=0(x-7)
:. -y +5= +0
:. +y-5=0
Second method :
Points are A(7,5),B(-9,5)
:. x_1=7,y_1=5,x_2=-9,y_2=5
Slope of the line, m=(y_2-y_1)/(x_2-x_1)
:. m=(5-5)/(-9-7)
:. m=(0)/(-16)
:. Slope =0
The equation of a line with slope m and passing through (x_1,y_1) is y-y_1=m(x-x_1)
Here Point (x_1,y_1)=(7,5) and Slope m=0 (given)
:. y-5=0(x-7)
:. y -5=0x +0
:. +y-5=0
Hence, The equation of line is +y-5=0
2. Find the equation of a straight line passing through the points A(-1,1) and B(2,-4)
Solution:
The given points are A(-1,1),B(2,-4)
:. x_1=-1,y_1=1,x_2=2,y_2=-4
Using two-points formula, The equation of a line AB is
(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)
:. (y-1)/(-4-1)=(x+1)/(2+1)
:. (y-1)/(-5)=(x+1)/(3)
:. 3(y-1)=-5(x+1)
:. 3y -3=-5x -5
:. 5x+3y+2=0
Second method :
Points are A(-1,1),B(2,-4)
:. x_1=-1,y_1=1,x_2=2,y_2=-4
Slope of the line, m=(y_2-y_1)/(x_2-x_1)
:. m=(-4-1)/(2+1)
:. m=(-5)/(3)
:. Slope =-5/3
The equation of a line with slope m and passing through (x_1,y_1) is y-y_1=m(x-x_1)
Here Point (x_1,y_1)=(-1,1) and Slope m=-5/3 (given)
:. y-1=-5/3(x+1)
:. 3(y-1)=-5(x+1)
:. 3y -3=-5x -5
:. 5x+3y+2=0
Hence, The equation of line is 5x+3y+2=0
3. Find the equation of a straight line passing through the points A(-5,-6) and B(3,10)
Solution:
The given points are A(-5,-6),B(3,10)
:. x_1=-5,y_1=-6,x_2=3,y_2=10
Using two-points formula, The equation of a line AB is
(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)
:. (y+6)/(10+6)=(x+5)/(3+5)
:. (y+6)/(16)=(x+5)/(8)
:. (y+6)/(2)=(x+5)/(1)
:. (y+6)=2(x+5)
:. y +6=2x +10
:. 2x-y+4=0
Second method :
Points are A(-5,-6),B(3,10)
:. x_1=-5,y_1=-6,x_2=3,y_2=10
Slope of the line, m=(y_2-y_1)/(x_2-x_1)
:. m=(10+6)/(3+5)
:. m=(16)/(8)
:. m=2
:. Slope =2
The equation of a line with slope m and passing through (x_1,y_1) is y-y_1=m(x-x_1)
Here Point (x_1,y_1)=(-5,-6) and Slope m=2 (given)
:. y+6=2(x+5)
:. y +6=2x +10
:. 2x-y+4=0
Hence, The equation of line is 2x-y+4=0
4. Find the equation of a straight line passing through the points A(3,-5) and B(4,-8)
Solution:
The given points are A(3,-5),B(4,-8)
:. x_1=3,y_1=-5,x_2=4,y_2=-8
Using two-points formula, The equation of a line AB is
(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)
:. (y+5)/(-8+5)=(x-3)/(4-3)
:. (y+5)/(-3)=(x-3)/(1)
:. (y+5)=-3(x-3)
:. y +5=-3x +9
:. 3x+y-4=0
Second method :
Points are A(3,-5),B(4,-8)
:. x_1=3,y_1=-5,x_2=4,y_2=-8
Slope of the line, m=(y_2-y_1)/(x_2-x_1)
:. m=(-8+5)/(4-3)
:. m=(-3)/(1)
:. m=-3
:. Slope =-3
The equation of a line with slope m and passing through (x_1,y_1) is y-y_1=m(x-x_1)
Here Point (x_1,y_1)=(3,-5) and Slope m=-3 (given)
:. y+5=-3(x-3)
:. y +5=-3x +9
:. 3x+y-4=0
Hence, The equation of line is 3x+y-4=0
5. Find the equation of a straight line passing through the points A(-1,-4) and B(3,0)
Solution:
The given points are A(-1,-4),B(3,0)
:. x_1=-1,y_1=-4,x_2=3,y_2=0
Using two-points formula, The equation of a line AB is
(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)
:. (y+4)/(-0+4)=(x+1)/(3+1)
:. (y+4)/(4)=(x+1)/(4)
:. (y+4)/(1)=(x+1)/(1)
:. (y+4)=(x+1)
:. y +4=x +1
:. x-y-3=0
Second method :
Points are A(-1,-4),B(3,0)
:. x_1=-1,y_1=-4,x_2=3,y_2=0
Slope of the line, m=(y_2-y_1)/(x_2-x_1)
:. m=(0+4)/(3+1)
:. m=(4)/(4)
:. m=1
:. Slope =1
The equation of a line with slope m and passing through (x_1,y_1) is y-y_1=m(x-x_1)
Here Point (x_1,y_1)=(-1,-4) and Slope m=1 (given)
:. y+4=1(x+1)
:. y +4=x +1
:. x-y-3=0
Hence, The equation of line is x-y-3=0
This material is intended as a summary. Use your textbook for detail explanation.
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