Equation of a line given slope and y intercept calculator (2024)
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Method and examples
Method
Equation of a line given slope and y intercept
1. Distance, Slope of two points
1. Find the distance between the points `A(5,-8)` and `B(-7,-3)`
2. Find the slope of the line joining points `A(4,-8)` and `B(5,-2)`
A(,),B(,)
`A(5,-8),B(-7,-3)`
`A(7,-4),B(-5,1)`
`A(-6,-4),B(9,-12)`
`A(1,-3),B(4,-6)`
`A(-5,7),B(-1,3)`
`A(-8,6),B(2,0)`
`A(0,0),B(7,4)`
Find the value of x or y
3. If distance between the points (5,3) and (x,-1) is 5, then find the value of x.
A(,),B(,),Distance =
`A(5,3),B(x,-1)`, distance `=5`
`A(x,-1),B(3,2)`, distance `=5`
`A(x,2),B(3,-6)`, distance `=10`
`A(x,1),B(-1,5)`, distance `=5`
`A(x,7),B(1,15)`, distance `=10`
`A(1,x),B(-3,5)`, distance `=5`
`A(x,0),B(4,8)`, distance `=10`
4. If slope of the line joining points `A(x,0), B(-3,-2)` is `2/7`, find the value of `x`
A(,),B(,),Slope =
`A(x,0),B(-3,-2)`, slope `=2/7`
`A(2,x),B(-3,7)`, slope `=1`
`A(x,5),B(-1,2)`, slope `=3/4`
`A(2,5),B(x,3)`, slope `=2`
`A(x,2),B(6,-8)`, slope `=-5/4`
`A(-2,x),B(5,-7)`, slope `=-1`
`A(2,3),B(x,6)`, slope `=3/5`
`A(-3,4),B(5,x)`, slope `=-5/4`
`A(0,x),B(5,-2)`, slope `=-9/5`
2. Points are Collinear or Triangle or Quadrilateral form
Show that the points are the vertices of
Find `A(0,0), B(2,2), C(0,4), D(-2,2)` are vertices of a square or not
A(,),B(,),C(,),D(,)
`A(1,5),B(2,3),C(-2,-11)` are collinear points
`A(1,-3),B(2,-5),C(-4,7)` are collinear points
`A(-1,-1),B(1,5),C(2,8)` are collinear points
`A(0,-1),B(3,5),C(5,9)` are collinear points
`A(2,8),B(1,5),C(0,2)` are collinear points
`A(-1,-1),B(1,5),C(2,8)` are collinear points
`A(0,-1),B(3,5),C(5,9)` are collinear points
`A(2,8),B(1,5),C(0,2)` are collinear points
`A(0,0),B(0,3),C(4,0)` are vertices of a right angle triangle
`A(-2,-2),B(-1,2),C(3,1)` are vertices of a right angle triangle
`A(-3,2),B(1,2),C(-3,5)` are vertices of a right angle triangle
`A(2,5),B(8,5),C(5,10.196152)` are vertices of an equilateral triangle
`A(2,2),B(-2,4),C(2,6)` are vertices of an isosceles triangle
`A(0,0),B(2,0),C(-4,0),D(-2,0)` are collinear points
`A(3,2),B(5,4),C(3,6),D(1,4)` are vertices of a square
`A(0,0),B(2,2),C(0,4),D(-2,2)` are vertices of a square
`A(1,-1),B(-2,2),C(4,8),D(7,5)` are vertices of a rectangle
`A(0,-4),B(6,2),C(3,5),D(-3,-1)` are vertices of a rectangle
`A(3,0),B(4,5),C(-1,4),D(-2,-1)` are vertices of a rhombus
`A(2,3),B(7,4),C(8,7),D(3,6)` are vertices of a parallelogram
`A(1,5),B(1,4),C(-1,3),D(-1,4)` are vertices of a parallelogram
3. Find Ratio of line joining AB and is divided by P
1. Find the ratio in which the point P(3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5)
P(,),A(,),B(,)
`P(3/4,5/12),A(1/2,3/2),B(2,-5)`
`P(-1,6),A(3,10),B(6,-8)`
`P(-2,3),A(-3,5),B(4,-9)`
`P(3,10),A(5,12),B(2,9)`
`P(6,17),A(1,-3),B(3,5)`
`P(12,23),A(2,8),B(6,14)`
`P(3,10),A(5,12),B(2,9)`
`P(6,17),A(1,-3),B(3,5)`
`P(12,23),A(2,8),B(6,14)`
`P(17/5,47/5),A(5,13),B(1,4)`
2. Write down the co-ordinates of the point P that divides the line joining A(-4,1) and B(17,10) in the ratio 1:2
A(,),B(,),ratio = : ,
`A(5,13),B(1,4),m:n=2:3`
`A(-4,1),B(17,10),m:n=1:2`
`A(5,12),B(2,9),m:n=2:1`
`A(2,8),B(6,14),m:n=5:3` Externally
`A(1,-3),B(3,5),m:n=5:3` Externally
3. In what ratio does the x-axis divide the join of `A(2,-3)` and `B(5,6)`? Also find the coordinates of the point of intersection.
A(,),B(,),divided by
`A(2,-3),B(5,6)` divided by x-axis
`A(1,2),B(2,3)` divided by x-axis
`A(5,-6),B(-1,-4)` divided by y-axis
`A(-2,1),B(4,5)` divided by y-axis
`A(2,1),B(7,6)` divided by x-axis
`A(2,-4),B(-3,6)` divided by y-axis
4. Find the ratio in which the point `P(x,4)` divides the line segment joining the points `A(2,1)` and `B(7,6)`? Also find the value of `x`.
P(,),A(,),B(,),
`P(x,2),A(12,5),B(4,-3)`
`P(11,y),A(15,5),B(9,20)`
`P(-3,y),A(-5,-4),B(-2,3)`
`P(-4,y),A(-6,10),B(3,-8)`
`P(x,4),A(2,1),B(7,6)`
`P(x,0),A(2,-4),B(-3,6)`
`P(0,y),A(2,-4),B(-3,6)`
4. Find Midpoint or Trisection points or equidistant points on X-Y axis
1. Find the coordinates of the midpoint of the line segment joining the points `A(-5,4)` and `B(7,-8)`
2. Find the trisectional points of line joining `A(-3,-5)` and `B(-6,-8)`
3. Find the point on the x-axis which is equidistant from `A(5,4)` and `B(-2,3)`
4. Find the point on the y-axis which is equidistant from `A(6,5)` and `B(-4,3)`
A(,),B(,)
`A(-5,4),B(7,-8)`
`A(2,1),B(1,-3)`
`A(2,1),B(5,3)`
`A(3,-5),B(1,1)`
`A(1,-1),B(-5,-3)`
`A(-7,-3),B(5,3)`
5. Find Centroid, Circumcenter, Area of a triangle
1. Find the centroid of a triangle whose vertices are `A(4,-6),B(3,-2),C(5,2)`
2. Find the circumcentre of a triangle whose vertices are `A(-2,-3),B(-1,0),C(7,-6)`
3. Using determinants, find the area of the triangle with vertices are `A(-3,5),B(3,-6),C(7, 2)`
4. Using determinants show that the following points are collinear `A(2,3),B(-1,-2),C(5,8)`
A(,),B(,),C(,)
`A(4,-6),B(3,-2),C(5,2)`
`A(3,-5),B(-7,4),C(10,-2)`
`A(4,-8),B(-9,7),C(8,13)`
`A(3,-7),B(-8,6),C(5,10)`
`A(2,4),B(6,4),C(2,0)`
6. Find the equation of a line using slope, point, X-intercept, Y-intercept
1. Find the equation of a straight line passing through `A(-4,5)` and having slope `-2/3`
A(,),Slope :
`A(-4,5)`,slope`=-2/3`
`A(4,5)`,slope`=1`
`A(-2,3)`,slope`=-4`
`A(-1,2)`,slope`=-5/4`
`A(0,3)`,slope`=2`
`A(0,0)`,slope`=1/4`
`A(5,4)`,slope`=1/2`
2. Find the equation of a straight line passing through the points `A(7,5)` and `B(-9,5)`
A(,),B(,)
`A(7,5),B(-9,5)`
`A(-1,1),B(2,-4)`
`A(-5,-6),B(3,10)`
`A(3,-5),B(4,-8)`
`A(-1,-4),B(3,0)`
`A(7,8),B(1,0)`
`A(6,4),B(-1,5)`
`A(2,3),B(7,6)`
`A(-3,4),B(5,-6)`
`A(0,7),B(5,-2)`
`A(0,0),B(-4,-6)`
`A(3,5),B(6,4)`
3. Find the equation of a line having slope `1/2` and y-intercept `-3`
Slope : ;Y-intercept :
slope`=1/2`,Y-intercept`=-3`
slope`=-1/2`,Y-intercept`=3`
slope`=2`,Y-intercept`=-3`
slope`=2`,Y-intercept`=3/2`
slope`=1/2`,Y-intercept`=5`
slope`=1/4`,Y-intercept`=0`
slope`=2`,Y-intercept`=3`
4. Find the equation of a line whose x-intercept is 5 and y-intercept is 2
X-intercept : ;Y-intercept :
X-intercept`=5`,Y-intercept`=2`
X-intercept`=3`,Y-intercept`=-2`
X-intercept`=-2/7`,Y-intercept`=3/5`
X-intercept`=6`,Y-intercept`=-4`
X-intercept`=2`,Y-intercept`=-2`
X-intercept`=-5/3`,Y-intercept`=5`
X-intercept`=-3/5`,Y-intercept`=-3/2`
X-intercept`=3`,Y-intercept`=-5`
7. Find Slope, X-intercept, Y-intercept of a line
1. Find the slope and y-intercept of the line 2x+3y=4
2. Find x-intercept and y-intercept of the line 2x+3y=4
3. Find the slope, x-intercept and y-intercept of the line 2x+3y=4
Line :
Line`:2x+3y=4`
Line`:2x+3y-6=0`
Line`:3x+6y-8=0`
Line`:4x+5y+7=0`
Line`:2x+3y+4=0`
Line`:3x+6y-8=0`
Line`:4x+5y+7=0`
Line`:3x-2y-12=0`
Line`:7y-4x+9=0`
Line`:5x+2y-11=0`
Line`:3x-y+11=0`
Line`:4x-3y+2=0`
Click here to Find Line using Slope, X-intercept, Y-intercept Calculator
4. Find the slope, x-intercept and y-intercept of the line joining the points `A(1,3)` and `B(3,5)`
A(,),B(,)
`A(1,3),B(3,5)`
`A(4,-8),B(5,-2)`
`A(7,1),B(8,9)`
`A(4,8),B(5,5)`
`A(7,8),B(1,0)`
`A(6,4),B(-1,5)`
`A(2,3),B(7,6)`
`A(-3,4),B(5,-6)`
`A(0,7),B(5,-2)`
`A(0,0),B(-4,-6)`
`A(3,5),B(6,4)`
`A(3,-5),B(-7,9)`
8. Find the equation of a line passing through point of intersection of two lines and slope or a point
1. Find the equation of a line passing through the point of intersection of lines `3x+4y=7` and `x-y+2=0` and having slope 5
Line-1 : ,
Line-2 : ,
Slope :
Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
2. Find the equation of a line passing through the point of intersection of lines `4x+5y+7=0` and `3x-2y-12=0` and point `A(3,1)`
Line-1 : ,
Line-2 : ,
A(,)
Line-1`:x+y+1=0`,Line-2`:3x+y-5=0`,`A(1,-3)`
Line-1`:4x+5y+7=0`,Line-2`:3x-2y-12=0`,`A(3,1)`
9. Find the equation of a line passing through a point and parallel or perpendicular to Line-2
1. Find the equation of the line passing through the point `A(5,4)` and parallel to the line `2x+3y+7=0`
2. Find the equation of the line passing through the point `A(1,1)` and perpendicular to the line `2x-3y+2=0`
A(,),Line-2 :
`A(5,4)`,Line`:2x+3y+7=0`
`A(1,1)`,Line`:2x-3y+2=0`
`A(2,3)`,Line`:2x-3y+8=0`
`A(2,-5)`,Line`:2x-3y-7=0`
3. Find the equation of the line passing through the point `A(1,3)` and parallel to line passing through the points `B(3,-5)` and `C(-6,1)`
4. Find the equation of the line passing through the point `A(5,5)` and perpendicular to the line passing through the points `B(1,-2)` and `C(-5,2)`
A(,),B(,),C(,),
`A(1,3),B(3,-5),C(-6,1)`
`A(4,-5),B(3,7),C(-2,4)`
`A(-1,3),B(0,2),C(4,5)`
`A(2,-3),B(1,2),C(-1,5)`
`A(4,2),B(1,-1),C(3,2)`
`A(5,5),B(1,-2),C(-5,2)`
10. Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3
1. Find the equation of the line passing through the point of intersection of the lines `x-y=1` and `2x-3y+1=0` and parallel to the line `3x+4y=12`
2. Find the equation of the line passing through the point of intersection of the lines `2x+3y=1` and `3x+4y=6` and perpendicular to the line `5x-2y=7`
Given two points on a line, we can write an equation for that line by finding the slope between those points, then solving for the y-intercept in the slope-intercept equation y=mx+b.
Given two points on a line, we can write an equation for that line by finding the slope between those points, then solving for the y-intercept in the slope-intercept equation y=mx+b.
These are the two methods to finding the equation of a line when given a point and the slope: Substitution method = plug in the slope and the (x, y) point values into y = mx + b, then solve for b. Use the m given in the problem, and the b that was just solved for, to create the equation y = mx + b.
We can rewrite an equation in slope-intercept form (y=mx+b) to be in standard form (Ax+By=C) instead. In this example, we rewrite the slope-intercept equation y=2/3x+4/7 in standard form.
Explanation: These lines are written in the form y = mx + b, where m is the slope and b is the y-intercept. We know from the question that our slope is 3 and our y-intercept is –5, so plugging these values in we get the equation of our line to be y = 3x – 5.
The equation is y = mx + b. The x and y variables remain as letters, but m and b are replaced by numbers (ex: y = 2x + 4, slope = 2 and y-intercept = 4).
What is the Slope Intercept Form of a Line? The graph of the linear equation y = mx + c is a line with m as slope, m and c as the y-intercept. This form of the linear equation is called the slope-intercept form, and the values of m and c are real numbers. The slope, m, represents the steepness of a line.
Given the graph of a line, you can determine the equation in two ways, using slope-intercept form, y=mx+b, or point-slope form, y−y1=m(x−x1). The slope and one point on the line is all that is needed to write the equation of a line. All nonvertical lines are completely determined by their y-intercept and slope.
What is average rate of change? It is a measure of how much the function changed per unit, on average, over that interval. It is derived from the slope of the straight line connecting the interval's endpoints on the function's graph.
A line has an undefined slope when it is a vertical line. A vertical line has no horizontal distance to it which is needed to have a positive, negative, or zero slope. An example of the equation of a line with an undefined slope would be x = 4 .
Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept). We can rewrite an equation in point-slope form to be in slope-intercept form y=mx+b, to highlight the same line's slope and y-intercept.
Steps to find the equation of a line from two points:
Find the slope using the slope formula. ...
Use the slope and one of the points to solve for the y-intercept (b). ...
Once you know the value for m and the value for b, you can plug these into the slope-intercept form of a line (y = mx + b) to get the equation for the line.
The letter m represents the slope of the line, or the rate at which y is changing with respect to x. The letter b represents the y-intercept of the line, or the y-coordinate of the point where the line crosses the y-axis.
How to Find the Equation of a Line from Two Points
Find the slope using the slope formula. ...
Use the slope and one of the points to solve for the y-intercept (b). ...
Once you know the value for m and the value for b, you can plug these into the slope-intercept form of a line (y = mx + b) to get the equation for the line.
Since c is the y-intercept of the line the line passes through the point 0 c and has slope m. Therefore the equation of the line isy - c = mx - 0⇒ y = mx + c.
The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis. The equation of a straight line with gradient m and intercept c on the y-axis is y = mx + c.
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