Second Term Lesson Note for Week One
Class : SSS 2
Subject : Mathematics
Topic : Revision on Lines (Gradient of lines)
Duration : 80 Minutes
Period : Double Period
Reference Book :
Instructional Material :
Learning Objectives : By the end of the lesson learners will be able to :
i. Define lines and Gradient of a line
ii. Calculate the gradients of a line.
Content :
REVISION/STRAIGHT LINE
GRADIENTS OF A STRAIGHT LINE AND GRADIENT OF A CURVE
In coordinate geometry, we make use of points in a plane. A point consists of the x-coordinate called abscissa and the y-coordinate known as ordinate. In locating a point on the x – y plane x – coordinate is first written and then the y-coordinate. For example, in a given point (a, b), the value of x is a and that of y is b. Similarly, in a point (3, 5), the value of x is 3 and that of y is 5. A linear graph gives a straight line graph from any given straight line equation which is in the general form y = mx + c or ax + by + c = 0
Example: Draw the graph of equation 4x + 2y = 5
Point of intersection of two linear equations
Two lines y = ax +b and y2 = cx + d
Intercept when ax + b = cx + d
That is you solve the two equations simultaneously
Intersection of a line with the x or y axis
The point of intersection of a line with the x –axis can be obtained by putting y = o to find the corresponding value of x = a, say the required point of intersection gives (a, o). Similarly, for the point of intersection of a line with the y-axis, put x = o to find the corresponding value of y. If the corresponding value of y is b, the required point of intersection is (o, b)
Example: Find the point of intersection of the line 2x + 3y + 2 = 0 with the
i. x – axis (ii) y – axis
Example 3: Find the point of intersection of the lines y = 3x + 2 and y = 2x + 5
Solution
y = 3x + 2 (1)
y = 2x + 5 (2)
At the point of intersection
3x + 2 = 2x + 5
3x – 2x = 5 -2
X = 3
Substitute 3 for x in equation (1), we obtain y = 3(3) + 2 = 11.
Hence, the point of intersection is (3, 11)
GRADIENT OF A STRAIGHT LINE
The Gradient of a straight line is defined as the ratio
Change in y in moving from one
Change in x point to another on the line. The Gradient of a straight line is always constant.
Gradient from A to B =
Gradient of a line that passes through points (x1, y1)
And (x2, y2) is given as gradient m =
Meaning that the gradient of the line is the ratio of increase in y to increase in x
TANGENT OF ANGLE OF SLOPE
In the above diagram, tan = .
Since = y2 – y1 tan = = m
And = x2 – x1
tan = m. it then follows that the gradient of a line can be defined as tangent of angle of slope.
Example:
Calculate the gradient and the angle of slope of the line passing through (1, 3) and (-4, 2)
EQUATION OF A STRAIGHT LINE AND TANGENT TO A CURVE
EQUATION OF A STRAIGHT LINE
Equation of a line with gradients in m and y intercept c. Equation of a line with gradient m and y intercept c is given as y = mx + c.
ii. Equation of a line passing through the point (x1, y1) with gradient.
The general equation of a line with known gradient m and which passes through the point (x1,y1) is given as m = –
Example 2
The equation of the line with gradient 2 and which passé through the point (-3, 2).
The solution (equation) of a line with known gradient and passing through the point (x1, y1) is given by –
Here, m = 2, (x1, y1) = (-3, 2)
The required equation of the line is
y – 2 = 2(x + 3)
y = 2x + 6 + 2
y = 2x + 8
y = 2x + 8
Equation of a line passing through two given points
The equation of a line passing through two given points (x1, y1) and (x2, y2)
Is =
iv. Double Intercept form of the equation of a line. The equation of a line which has an intercept “a” on the x – axis and intercept “b” on the y-axis is given by + = 1
Double intercept form of the equation of a line
The Equation of a line which has an intercept ‘a’ on the axis and intercept ‘b’ on the y-axis is given by + = 1
v. Equation of a line passing through a point and making an angle with the horizontal axis.
The equation of a line passing through the point (x1, y1) and making an angle with the horizontal axis is = tan or y – y1 = (x – x1) tan .
Drawing Tangents to a cure
The gradient at any particular point on a curve is defined as being the gradient of the tangent to the curve at that point the gradient of the curve at point A is the gradient of the tangent BA, that is, tan . The tangent is drawn by placing a ruler against the curve at A and drawing a line considering that the angels between the line and the curve are equal. (Note: Gradient to the horizontal line of a curve is zero because the tangent is horizontal known as a turning points (maximum/minimum)
WRAP UP AND ASSESSMENT
The gradients of a straight line is given as gradient = change in y / change in x.
The gradient of a curve at a point is given by the gradient of the tangent at that point.
The gradient at a turning point of any quadratic equation equals zero.
Exercise 14.5 no 1; the figure below (the text recommended) represents the graph of the function y = x2 + 4x – 5, (a) use the given tangents to find the gradient of the curve at (i) A (ii) B.
(b) Use the Graph to find the roots of the function.
(c) State the equation of the line of symmetry of the curve.
2. Find the value of h if the line joining (h, 2) and (3, 1/3) has gradient of – 2/3.
3. Find the distance and midpoint between the points K(-1, 6) and Q(3,- 4).
4. If the distance between (1, 2) and (5, y) is 5 units, find y.
5. If M(4, p) is the midpoint of the line joining L(p, – 2) and N(q, p). Find the values of p and q.
ASSIGNMENT :
Ex 14.5 Pg 194, No 2. And 4 copy and complete the table below for the function y = use a scale of 2cm for 1 unit on both axes, draw the graph of the function.
2. Find the point of intersection of the two lines 3x – 2y + 5 = 0 and y – 4x + 3 = 0
3. Find the equation of the line of point which is equivalent from the point (2, 5) and (3, 4).
4. If the coordinate of the midpoint of the line joining (3, x) and (y, 10) are (1, 3). Find the values of x and y.
5. Find the distance between the point Q(4, 3) and the point common to the lines 2x – y – 4 and x + y = 2.