Analytical Geometry Grade 10
1. Finding the equation of a straight line, you could either be given:
a. Two points A (Xa; Xb) and B (Xb; Xb) – Find the gradient using gradient formula which is
𝑌𝑏−𝑌𝑎
M= 𝑋𝑏−𝑋𝑎
.
- After finding the gradient, update the equation by
substituting the value of m in the formula and find
the value of c by either substituting any point
strictly on that line and solve for the value of c.
Example: Find the equation of a line passing through the points A (8; 4) and B (6; 2).
Sol: Step 1- Remember the general equation of a straight line is y= mx + c. So, in order to from the
equation we need the value of m and c.
𝑌𝑏−𝑌𝑎 2−4 −2 1
Step 2 – We find the value of the gradient using the gradient formula i.e., m = 𝑋𝑏−𝑋𝑎 = 6−8 = −4= 2
Step 3- We update the equation by putting our answer where we see (m) in our equation, so our
1
equation now becomes Y = 𝑥 + C. We now need the value of c for it to be complete. So we find C by
2
substituting: It can be any point as long as it lies on the line, in our case we can choose point A (8; 4)
so,
1
Y= 4 and X=8 so we have 4 = (8) + c
2
4=4+c
4-4 = c
1 1
Therefore, our value of c is 0. We can update our equation by putting in the value of c. Y= 2x + 0 = 2 𝑥
2. You can be given a point and get told that the line going through that point is parallel to a line y
= mx + c.
Example: AB is parallel to line with equation y = 6x + 4 and goes through the point of A ( 2; 0).
Find the equation of line AB.
Sol: Remember the equation of a straight line is always in the form of Y = mx + c and as always,
we need the value of m and c to have the complete equation.
Step 1: They told us the two lines are PARALLEL and according to what we know, parallel lines
have equal gradients, meaning the value of m= 6. Then, we can update our equation by
substituting the value of m. Our equation becomes Y= 6x + C.
Step 2: Now we need the value of C. We can find c by substituting the point A(2;0) and solving
for c i.e 0 = 6(2) + C
0 = 12 + C
-12 = C
Therefore equation of line AB is Y = 6x -12.
1. Finding the equation of a straight line, you could either be given:
a. Two points A (Xa; Xb) and B (Xb; Xb) – Find the gradient using gradient formula which is
𝑌𝑏−𝑌𝑎
M= 𝑋𝑏−𝑋𝑎
.
- After finding the gradient, update the equation by
substituting the value of m in the formula and find
the value of c by either substituting any point
strictly on that line and solve for the value of c.
Example: Find the equation of a line passing through the points A (8; 4) and B (6; 2).
Sol: Step 1- Remember the general equation of a straight line is y= mx + c. So, in order to from the
equation we need the value of m and c.
𝑌𝑏−𝑌𝑎 2−4 −2 1
Step 2 – We find the value of the gradient using the gradient formula i.e., m = 𝑋𝑏−𝑋𝑎 = 6−8 = −4= 2
Step 3- We update the equation by putting our answer where we see (m) in our equation, so our
1
equation now becomes Y = 𝑥 + C. We now need the value of c for it to be complete. So we find C by
2
substituting: It can be any point as long as it lies on the line, in our case we can choose point A (8; 4)
so,
1
Y= 4 and X=8 so we have 4 = (8) + c
2
4=4+c
4-4 = c
1 1
Therefore, our value of c is 0. We can update our equation by putting in the value of c. Y= 2x + 0 = 2 𝑥
2. You can be given a point and get told that the line going through that point is parallel to a line y
= mx + c.
Example: AB is parallel to line with equation y = 6x + 4 and goes through the point of A ( 2; 0).
Find the equation of line AB.
Sol: Remember the equation of a straight line is always in the form of Y = mx + c and as always,
we need the value of m and c to have the complete equation.
Step 1: They told us the two lines are PARALLEL and according to what we know, parallel lines
have equal gradients, meaning the value of m= 6. Then, we can update our equation by
substituting the value of m. Our equation becomes Y= 6x + C.
Step 2: Now we need the value of C. We can find c by substituting the point A(2;0) and solving
for c i.e 0 = 6(2) + C
0 = 12 + C
-12 = C
Therefore equation of line AB is Y = 6x -12.
