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Point of intersection formula is used to find the point of intersection of two lines, which means the meeting point of two lines. The point of intersection of two curves is the point where two curves take the same value. This concept of point of intersection can be useful in a variety of applications. The intersection of lines can be an empty set, a point or a line in Euclidean geometry. The two lines can be represented by the equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. It can be possible to find the point of intersection of three or more lines. By solving the above two equations, we can find the solution for the point of intersection of two lines.
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Key Takeaways: Point Gradient, Intersection, points, gradient
Point of Intersection Formula
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Point of intersection is defined as the point where two lines or two curves meet with each other. In simple words, it is defined as the common point of both lines and curves that satisfies both the curves which can be derived by solving the equation of curves. Consider 2 straight lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 which intersects at a point (x, y). So, we need to find a line intersection formula to find these points of intersection(x, y). These points should satisfy both the equations. By solving the two equations, we can find the point where two lines intersect. The formula for point of intersection of two lines:
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(x, y) = [\(\frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} , \frac{a_2c_1 - a_1c_2}{a_1b_2 - a_2b_1} \)]Point of Intersection formula
The video below explains this:
Coordinate Geometry Detailed Video Explanation:
Derivation of Point of Intersection Formula
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Given two equations,
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Solving the equations using cross multiplication method:
x y 1
b1 c1 a1 b1
b2 c2 a2 b2
On cross multiplying both the equations, we will get:
x/(b1*c2-b2*c1) = y/(c1*a2-c2*a1) = 1/(a1*b2-a2*b1)
Now solving for x we get,
x/(b1*c2-b2*c1) = 1/(a1*b2-a2*b1)
x = (b1*c2-b2*c1)/(a1*b2-a2*b1)
Now solving for y we get,
y/(c1*a2-c2*a1) = 1/(a1*b2-a2*b1)
x = (c1*a2-c2*a1)/(a1*b2-a2*b1)
Hence, the formula for point of intersection will be:
(x, y) = ((b1*c2-b2*c1)/(a1*b2-a2*b1), (c1*a2-c2*a1)/(a1*b2-a2*b1))
If the line segments are parallel, then they will never intersect with each other.
Condition of two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 to be parallel is a1/b1 = a2/b2
Point of Intersection
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Point Gradient
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In a line, point gradient is defined as the ratio of vertical change to the horizontal change through a point. In other words, the gradient of a line will be rise divided by run i.e rise/run. If m is the gradient point across a line segment, the point gradient formula will be:
Point Gradient = \(\frac{y-y_1}{x - x_1}\)
Point Gradient Formula
Point Slope Form Formula
Another popular format for straight line equations is the point slope formula. For its calculation, we need to find out the values of (x1, y1) and slope m.
y – y1 = m (x – x1)
Point Slope Formula
Where, m is known as the slope of line segment
x1 is the coordinate of x-axis.
y1 is the coordinate of the y-axis.
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Things to Remember
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- A point of intersection is defined as the meeting point of two lines or curves at a particular point. We can find a point of intersection in both ways i.e algebraically and graphically in a line segment.
- By drawing two curves on the same graph and finding their points of intersection, we can find a point of junction graphically.
- The maximum number of intersecting points are three if we have drawn three lines on a plane. When only one line is drawn, then there is only one intersection point. Hence, the maximum number of intersections of two lines is 1. The new line will only be able to intersect with the two old lines and a given pair of two lines can intersect at most once with each other.
- Some common examples of points of intersection in real life are a pair of scissors, a folding chair, a road cross, a signboard etc.
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Sample Questions
Ques. Check whether two lines are parallel or not 2x+4y+6 = 0 and 4x+8y+6 = 0. (2 Marks)
Ans. To check whether two lines are parallel or not, we should check a1/b1 = a2/b2
In this case, a1=2, b1=4
a2=4, b2=8
2/4 = 4/8
1/2 = 1/2
As the condition is satisfied, the lines are parallel and will not intersect with each other.
Ques. If λx2-10xy+12y2+5x-16y-3 = 0 represents a pair of straight lines, then find the value of λ. (2 Marks)
Ans. Given equation, λx2-10xy+12y2+5x-16y-3 = 0
Compare it with the standard eq: ax2-2hxy+by2+2gx-2fy+c = 0
Here, a = λ, b = 12, c = -3, f = -8, g = 5/2, h = -5
Using condition abc+2fgh-af2-bg2-ch2 = 0, we have
λ(12)(-3)+2(-8)(5/2)(-5)-λ(64)-12(25/4)+3(25) = 0
-36λ+200-64λ-75+75 = 0
100λ = 200
λ = 2
Hence, the value of λ comes out to be 2.
Ques. Check whether two lines are parallel or not 3x+4y+8 = 0 and 4x+8y+6 = 0. (2 Marks)
Ans. To check whether two lines are parallel or not, we should check a1/b1 = a2/b2
In this case, a1=3, b1=4
a2=4, b2=8
3/4 ≠ 4/8
As the condition is not satisfied, the lines are not parallel and will intersect with each other at any point.
Ques. Find the coordinates of the point of intersection of the lines 2x-y+3 = 0 and x+2y-4 = 0. (2 Marks)
Ans. Solving the equations 2x-y+3 = 0 and x+2y-4 = 0 simultaneously, we get
x/4-6 = y/3+8 = 1/4+1
x/-2 = y/11 = 1/5
x = -2/5, y = 11/5
Hence, (-2/5, 11/5) is the required point of intersection.
Ques. Check whether the point (3, 5) is the point of intersection of lines 2x+3y-21 = 0, x+2y-13 = 0. (3 Marks)
Ans. A point to be a point of intersection, it should satisfy both the lines.
Substituting (x, y) = (3, 5) in both the lines.
Checking for equation 1: 2*3+3*5-21 = 0
6+15-21 = 0
0 = 0 (satisfied for eq1)
Checking for equation 2: 3+2*5-13 = 0
3+10-13 = 0
0 = 0 (satisfied for eq2)
Hence, both the equations are satisfied (3, 5) is the point of intersection of both lines.
Ques. If x+4y-5 = 0 and 4x+ky+7 = 0 are two perpendicular lines then find the value of k. (3 Marks)
Ans. Find the slopes of both the equations
m1 = x+4y-5 = 0….(1)
y = mx+c….(2)
On comparing eq(1) and eq(2), we get
m1 = -1/4
m2 = 4x+ky+7 = 0….(3)
y = mx+c….(4)
On comparing eq(3) and eq(4), we get
m2 = -4/k
Two lines will be perpendicular to each other if m1m2 = -1
(-1/4)*(-4/k) = -1
k = -1
Hence, the value of k comes out to be -1.
Ques. Using the point of intersection formula, find the point of intersection of two lines 2x+4y+2 = 0 and 2x+3y+5 = 0. (3 Marks)
Ans. Given straight line equations are:
2x+4y+2 = 0 and 2x+3y+5 = 0
a1=2, b1=4, c1=2
a2=2, b2=3, c2=5
Intersection point can be find by the following formula:
(x, y) = ((b1*c2-b2*c1)/(a1*b2-a2*b1), (c1*a2-c2*a1)/(a1*b2-a2*b1))
(x, y) = ((4*5-3*2)/(2*3-2*4), (2*2-2*5)/(2*3-2*4))
(x, y) = ((20-6)/(6-8), (4-10)/(6-8))
(x, y) = (-7, 3)
Hence, the point of intersection is (-7, 3).
Ques. Check whether the point (2, 5) is the point of intersection of lines x+3y-17 = 0, x+y-13 = 0. (3 Marks)
Ans. A point to be a point of intersection, it should satisfy both the lines.
Substituting (x, y) = (2, 5) in both the lines.
Checking for equation 1: 2+3*5-17 = 0
2+15-17 = 0
0 = 0 (satisfied for eq1)
Checking for equation 2: 1*2+1*5-13 = 0
2+5-13 = 0
-6 = 0 (not satisfied for eq2)
Hence, The equation 2 is not satisfied (3, 5) is not the point of intersection of both lines.
Ques. What are the steps to find a point of intersection algebraically? (3 Marks)
Ans. Following are the steps to find a point of intersection algebraically:
- For any variable, let it be y solve for each equation.
- Set the equations for y in the first step to the same value then solve it for the other variable which will be x. This will be called the x-value of the junction point.
- In any of the original equations, put the value of x of the site of intersection and solve for y. This will be called the point of intersection of variable y.
Ques. Ques. Find the equation of line joining the point (3, 5) to the point of intersection of the lines 4x+y-1 = 0 and 7x-3y-35 = 0. (5 Marks)
Ans. Given straight line equations are:
4x+y-1=0 and 7x-3y-35=0
a1=4, b1=1, c1=-1
a2=7, b2=-3, c2=-35
Intersection point can be find by the following formula:
(x, y) = ((b1*c2-b2*c1)/(a1*b2-a2*b1), (c1*a2-c2*a1)/(a1*b2-a2*b1))
(x, y) = ((1*35-(-3)*(-1))/(4*3-7*1), (-1*7-(-35)*4)/(4*(-3)-7*1))
(x, y) = ((35+3)/(12+7), (-7+140)/(-12-7))
(x, y) = (2, -7)
So, the given lines intersect at (2, -7)
Now the equation of line joining the point (3, 5) and (2, -7) is
y-5 = -7-5 (x-3)/(2-3)
y-5 = 12x-36
12x-y-31 = 0 is the required equation of the line.
Ques. Find the equation of line parallel to the y-axis and draw through the point of intersection of lines x–7y+5 = 0 and 3x+y = 0. (5 Marks)
Ans. Given straight line equations are:
x–7y+5 = 0 and 3x+y = 0
a1=1, b1=-7, c1=5
a2=3, b2=1, c2=0
Intersection point can be find by the following formula:
(x, y) = ((b1*c2-b2*c1)/(a1*b2-a2*b1), (c1*a2-c2*a1)/(a1*b2-a2*b1))
(x, y) = ((-7*0-1*5)/(1*1-3*(-7)), (5*3-0*1)/(1*1-3*(-7)))
(x, y) = ((-5)/(22), (15)/(22))
(x, y) = (-5/22, 15/22)
Hence, the point of intersection is (-5/22, 15/22).
Let the equation of the required line be x = λ as the equation of a line parallel to y-axis is x = constant.
Since, equation passes through (-5/22, 15/22)
λ = -5/22
Substituting the value of λ in the above equation:
x = -5/22 or 22x+5 = 0
Therefore, 22x+5 = 0 is the equation of the required line segment.
Ques. Find the intersection point of the straight lines 3x+5y-6 = 0 and 5x-y-10 = 0 without using the standard formula.. (5 Marks)
Ans. Given straight line equations are:
3x+5y-6 = 0 and 5x-y-10 = 0
We have to subtract the following equations to get the values of x and y, in order to achieve that we should make a variable the same in both the equations.
Multiplying eq2 with 5, we will get
25x-5y-50 = 0….(1)
3x+5y-6 = 0….(2)
Solving (1) and (2)
28x-56 = 0
28x = 56
x = 2
Putting the value of x in eq(1) or (2) to get value of y,
3x-5y+6 = 0
3(2)-5y-6 = 0
6-5y-6 = 0
5y = 0
y = 0
So, The point of intersection of the straight lines 3x+5y-6 = 0 and 5x-y-10 = 0 will be (2, 0).
Ques. Find the intersection point of the straight lines 5x+6y = 25 and 3x-4y = 10 without using the standard formula. (5 Marks)
Ans. Given straight line equations are:
5x+6y = 25 and 3x-4y = 10
We have to subtract the following equations to get the values of x and y, in order to achieve that we should make a variable the same in both the equations.
Multiplying eq1 with 2
Multiplying eq2 with 3, we will get
10x+12y = 50
9x-12y = 30
Solving (1) and (2)
19x = 80
x = 80/19
Putting the value of x in eq(1) or (2) to get value of y,
5x+6y = 25
5(80/19)+6y = 25
400/19+6y = 25
(400+114y)/19 = 25
400+114y = 25*19
400+114y = 475
114y = 475-400
114y = 75
y = 25/38
So, The point of intersection of the straight lines 5x+6y = 25 and 3x-4y = 10 will be (80/19, 25/38).
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