Worksheet on Two Point Form helps students to understand the concept deeply. Get to know how to find the equation of a line in two-point form to solve Two-point Form Questions PDF. You can try to solve the given problems on this page and gain knowledge quickly. So, Check all the Two Point Form Examples with Solutions from the 10th Grade Math Worksheet as often as possible to understand the concept easily. Practice our Two-Point Form Worksheet PDF as many times as you wish to get a grip on all the problems.
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Finding the Equation of a Line Given Two Points Worksheet PDF
Problem 1: Find the Equation of the straight line passing through the points (2, 4), and (1, -2)?
Solution:
In the given question, the values
The points (x1, y1)is (2, 4) and (x2, y2) = (1, -2).
We know the formula. So, the equation for the straight line is (y – y1) = (y1– y2) /(x1– x2) * (x – x1).
Now, place the values in the above equation, then we will get
(y – 4) = (4 – (-2)) / (2 – 1) * (x – 2).
(y – 4) = (4 + 2) / (1) * (x – 2).
(y – 4) = 6 / (1) * (x – 2).
(y – 4) = 6 * ( x – 2).
y – 4 = 6x – 12.
6x + y – 4 -12 = 0.
6x + y – 16 = 0.
Therefore, the required straight line equation is 6x + y – 16 = 0.
Problem 2: Find the Equation of the straight line passing through the points (4, -4), and (3, 0)?
Solution:
In the given question, the values
The points (x1, y1)is (4, -4) and (x2, y2) = (3, 0).
We know the formula of the equation of a straight line is (y – y1) = (y1– y2) /(x1– x2) * (x – x1).
Now, replace the all given values in the above formula. Now, it will be,
(y – (-4)) = (-4 – 0) / (4 – 3) * (x – 4).
(y+4) = (-4) / (1) * (x – 4).
(y +4) = -4 / (1) * (x – 4).
(y + 4) = -4 * ( x – 4).
y + 4 = -4x +16.
4x + y + 4 -16 = 0.
4x + y – 12 = 0.
Therefore, the equation of a straight line is 4x + y – 12 = 0.
Problem 3: Find the Equation of the straight line passing through the points (a, 0), and (0, b)?
Solution:
As given in the question, the values
The points (x1, y1) is (a, 0) and
The points (x2, y2) is (0, b).
The formula for equation of a straight line in two-point form is (y – y1) = (y1– y2) /(x1– x2) * (x – x1).
Now, substitute the values in the two-point form formula.
Then the value is (y – 0) = (0 – b) / (a – 0) * (x – a).
(y – 0) = (-b) / (a) * (x – a).
y -0 = -b/(a)*(x – a).
y = -bx/a + ba/a
y= (-bx +ba)/a
ay = -bx +ba
ay+bx = ab
Now, dividing both sides by ab, then it will be
(y/b) + (x/a) = 1
So, the equation of a given line is, (x/a) + (y/b) = 1
Thus, the required straight line equation is (x/a)+(y/b) = 1.
Problem 4: Two different buildings’ heights are located on opposite sides of each other. If a heavy rod is attached to joining the terrace of the buildings from (4, 9) to (12, 16). What is the equation of the rod joining between the buildings?
Solution:
Given that,
The line passing through the points (x1, y1) is (4, 9) and
The points (x2, y2) are (12, 16).
Now, we will find the value of the equation of the rod joining the buildings.
So, the formula for equation of the line in two-point form is, (y – y1)/(y2Â – y1) = (x – x1)/(x2 – x1)
Now, we will place the given values in the above formula.
We have the values of (x1, y1) is (4, 9) and (x2, y2) is (12, 16).
After substitute, it will be
(y – 9)/(16 – 9) = (x – 4)/(12 – 4)
(y – 9)/7= (x – 4)/8
Next, find the LCM of the denominators.
So, the Least common multiple of the denominators 7 and 8 is 56.
Multiply each side by 56.
8(y – 10) = (x – 6)7
8y – 80 = 7x – 42
7x – 8y -80 +42 = 0
7x-8y-38=0
Hence, the required equation of the rod between the buildings is 7x – 8y -38 = 0.
Problem 5: What is the equation of a line which passes through the points (-1, 3) and (3,-2)?
Solution:
Given that,
The points (x1, y1) is (-1, 3) and the points (x2, y2) is (3, -2).
Now, we need to find the equation of a line with a two-point form.
The formula for two-point form is, (y – y1) = (y1– y2) /(x1– x2) * (x – x1).
Now, put the given (x1, y1) and (x2, y2) value in the formula. It will be,
(y – 3) = (3– (-2)) /((-1)– 3) * (x – (-1)).
(y-3) = (3+5)/(-1-3) *(x+1)
(y-3) = (8)/(-4)*(x+1)
(y-3) = (-2)*(x+1)
y-3 = -2x -2
2x+y-3-2 = 0
2x+y -5 =0
Hence, the required Equation of a line is 2x+y-5 =0.
Problem 6: Find the values of k for which the points A(k, -3), B(6, 3), and C(4, 2) are collinear?
Solution:
In the question the given information is,
The points A(k, -3), B(6, 3), and C(4, 2) are collinear.
As per formula, the values are (x, y) = (k, -3), (x1, y1) = (6, 3), and (x2, y2) = (4, 2)
We all know, the slope of the AB = slope of the BC.
The formula is (y1 – y) / (x1 – x) = (y2 – y1) / ((x2 – x1).
Substitute the values in the above equation. Then we will get
(3 – (-3)) / ( 6 – k) = (2 – 3) / (4 – 6).
(3 + 3) / (6 – k) = -1 / -2.
6 / (6 – k) = 1/2.
6 x 2 = 1(6 – k).
12 = 6-k
12 -6 = -k
k = -6
Hence, in the given points the value of k is -6.
Problem 7: Find the equation of a line in a slope-intercept form that passes through the points (3, 6) and (-2, 5)?
Solution:
Given that,
The points (x1, y1) is (3, 6).
The points x-coordinate and y-coordinate (x2, y2) is (-2, 5).
The formula for Equation of line in two-point form is (y – y1)/(y2 – y1) = (x – x1)/(x2 – x1).
Now, substitute (x1 , y1) and (x2, y2) values in a formula. We get,
(y – 6)/(5 – 6) = (x – 3)/(-2-3)
(y-6)/(-1) = (x-3)/(-5)
-5(y-6) = (-1)(x-3)
-5y +30 = -x + 3
x-5y+30-3=0
x-5y+27 =0
Hence, the required equation of a line is x-5y+27=0.
Problem 8: Using the two-point form method, find the equation of the line that passes through the points (-1,0) and (3,2).
Solution:
As given in the question,
The points (x1, y1) is (-1, 0) and the points (x2, y2) is (3, 2).
Now, we need to find the equation of a line with a two-point form.
The formula for two-point form is, (y – y1) = (y1– y2) /(x1– x2) * (x – x1).
Now, put the given (x1, y1) and (x2, y2) value in the formula. It will be,
(y – 0) = (0– 2) /((-1)– 3) * (x – (-1)).
(y-0) = (-2)/(-4) *(x+1)
(y-0) = (1)/(2)*(x+1)
(2)y = (1)*(x+1)
2y = x +1
-x+2y-1 = 0
-x+2y -1 =0
Thus, the required Equation of a straight line is -x+2y-1 =0.
Problem 9: Find the equation of a straight line with the help of two points (8, 4) and (6, 2)?
Solution:
As per the given information, the straight line crosses the two points.
The points with x-coordinate and y-coordinate is (x1, y1) = (8, 4) and (x2, y2) = (6, 2).
The equation for the straight line in Two-Point form is (y – y1) = (y1 – y2) / (x1 – x2) * (x – x1).
Replace the given values in the above equation, then it will be,
(y – 4) = (4 – 2) / (8 – 6) * (x – 8).
(y – 4) = (2) / (2) * (x – 8).
(y – 4) = (x – 8).
-x + y – 4 + 8 = 0.
-x + y + 4 = 0.
-x – y + 4 = 0.
Thus, the equation of a straight line in a Two–Points Form is -x-y+4 = 0.
Problem 10: Find the equation of the straight line which passes through the points (at1², 2at1) and (at2², 2at2)?
Solution:
In the given question,
The points on x-coordinate and y-coordinate is (x1, y1) = (at1², 2at1).
The points on (x2, y2) is (at2², 2at2).
Now, we will find out the equation of a line using the given points.
We know that, the equation of a line in two-point form formula is, (y – y1) = (y1 – y2) / (x1 – x2) * (x – x1).
Now, place the points values in the formula. Then we get,
(y – 2at1) = (2at1 – 2at2) / (at1²- at2²) * (x – at1²)
Now, simplify it.
(y – 2at1) = 2a(t1– t2)/ a(t1²- t2²) * (x – at1²)
2a(t1– t2)/ a(t1²- t2²)  is in the form of (a²- b²). The formula for (a²- b²) is (a-b)(a+b).
Now, the equation is (y – 2at1)= 2a(t1– t2)/ a(t1– t2)(t1 + t2) * a(t1– t2).
(y – 2at1)= 2/(t1 + t2) * (x – at1²)
(t1 + t2)(y – 2at1) =2(x-at1²)
yt1– 2at1²+ yt2– 2at1t2 = 2x – 2at1²
yt1– 2at1²+ yt2– 2at1t2 – 2x + 2at1² =0
– 2x +yt1 +yt2 -2at1²– 2at1t2 + 2at1² =0
– 2x +yt1 +yt2 – 2at1t2 =0
-2x + y(t1 + t2) – 2at1t2 =0
Therefore, the required straight line equation in two-point slope form is -2x + y(t1 + t2) – 2at1t2 =0.