*The straight line, L1, has an equation y= -2x+5
1.) Write down the gradient of L1
2.) Line 2, is perpendicular to L1, and passes through the point (4,5).
a. Write down the gradient of L2
b. Find the equation of L2.
3.) Line 2, is perpendicular to L1, and passes through the point (4,5). Write down the coordinatesof the point of intersection of L1 and L2.
gradient = slope
y=mx+b is called the slope-intercept for a reason
perpendicular lines have slopes whose product is -1
so, your new line has slope -1/m and its equation in point-slope form is
y-5 = -1/m (x-4)
Now work your magic. Come back with what you have if you get stuck.
But first, review the topic some more ...
1.) The gradient of L1 is -2.
2.)
a. The gradient of L2 can be found by taking the negative reciprocal of the gradient of L1. So, the gradient of L2 is 1/2.
b. To find the equation of L2, we can use the point-slope form of a line. Plugging in the gradient (1/2) and the point (4,5), we get: y - 5 = (1/2)(x - 4). Simplifying this, we get y = (1/2)x + 3.
3.) To find the point of intersection of L1 and L2, we can set their equations equal to each other and solve for x and y: -2x + 5 = (1/2)x + 3. Simplifying this, we get -2.5x = -2, and solving for x, we find x = 0.8. Plugging this back into L1's equation, we can find that y = 3.4. Therefore, the coordinates of the point of intersection are (0.8, 3.4).
1.) The gradient of L1 can be determined by comparing the equation with the form y = mx + c, where m represents the gradient. From the given equation y = -2x + 5, it can be seen that the coefficient of x is -2. Thus, the gradient of L1 is -2.
2.)
a. Since line 2 is perpendicular to L1, the product of the gradients of the two lines is -1. Therefore, the gradient of L2 will be the negative reciprocal of the gradient of L1. The negative reciprocal of -2 is 1/2.
b. To find the equation of L2, we can use the point-slope form of a linear equation: y - y1 = m(x - x1), where m is the gradient and (x1, y1) is any point on the line. Using the given point (4,5) and the gradient 1/2:
y - 5 = (1/2)(x - 4)
Simplifying the equation, we get:
y - 5 = 1/2x - 2
y = 1/2x + 3
Therefore, the equation of line 2 (L2) is y = 1/2x + 3.
3.) To find the point of intersection between L1 and L2, we can solve their equations simultaneously. Since we have the equations for both lines, we can set them equal to each other:
-2x + 5 = 1/2x + 3
To solve for x, we can simplify the equation:
-2x - 1/2x = 3 - 5
-4x/2 - x/2 = -2
(-4x - x) / 2 = -2
-5x / 2 = -2
-5x = -4 * 2
-5x = -8
x = -8 / -5
x = 8/5
Substituting the value of x back into either equation, we can find the corresponding y-coordinate. Let's use the equation of L1:
y = -2(8/5) + 5
y = -16/5 + 25/5
y = 9/5
Therefore, the point of intersection between L1 and L2 is (8/5, 9/5).
1.) To find the gradient (slope) of a straight line, you need to know its equation in the form y = mx + c. In the equation y = -2x + 5, the coefficient of x (-2) represents the gradient. Therefore, the gradient of L1 is -2.
2a.) Since Line 2 is perpendicular to L1, the gradient of L2 can be found by taking the negative reciprocal of the gradient of L1. In other words, the gradient of L2 will be the negative reciprocal of -2, which is 1/2.
2b.) We have the point (4,5) through which Line 2 passes, and we also know its gradient is 1/2. To find the equation of Line 2, we can use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is its gradient. Plugging in the values, we get:
y - 5 = (1/2)(x - 4)
Now, let's simplify this equation:
y - 5 = (1/2)(x) - (1/2)(4)
y - 5 = (1/2)x - 2
Bringing the constant terms to the right side, we get:
y = (1/2)x - 2 + 5
y = (1/2)x + 3
Therefore, the equation of Line 2 is y = (1/2)x + 3.
3.) To find the point of intersection between L1 and L2, we need to solve their equations simultaneously. The equations are:
L1: y = -2x + 5 (equation given)
L2: y = (1/2)x + 3 (equation derived in part 2b)
Now, we can set the two equations equal to each other and solve for x:
-2x + 5 = (1/2)x + 3
To simplify this equation, let's bring the x terms to one side:
-2x - (1/2)x = 3 - 5
(-4/2)x - (1/2)x = -2
-(5/2)x = -2
Next, we can isolate x by multiplying both sides of the equation by -2/5:
(-(5/2)x)(-2/5) = -2(-2/5)
x = 4/5
Now that we know x, we can substitute it back into either equation (L1 or L2) to find the corresponding y-coordinate. Let's use L1:
y = -2(4/5) + 5
y = -8/5 + 25/5
y = 17/5
Therefore, the point of intersection between L1 and L2 is (4/5, 17/5).