Math, asked by dgupta11, 1 month ago

Consider the line L with equation y + 2x = 3. The line L1

is parallel to L and passes through the

point (6, –4).
(a) Find the gradient of L1
.
(b) Find the equation of L1

in the form y = mx + b.

(c) Find the x-coordinate of the point where line L1 crosses the x-axis.

Answers

Answered by cromnium
1

Answer:

(a) The gradient ( which is I assume the slope) is -2.

(b) The equation of L1 is y = -2x + 8.

(c) The co ordinates are (4,0).

Step-by-step explanation:

(a) As we know that L and L1 are parallel to each other, the equation for L is y + 2x = 3. And the slope for this equation is -2 as y = mx + c.

y = -2x + 3.

(b) The equation of L1 can be found using the formula y-y1 = m(x-x1).

(y+4) = -2(x -6)

y +4 = -2x + 12

y = -2x + 8.

(c) Substitute 0 for y as the x axis co ordinate is (x,0).

0 = -2x + 8

2x = 8

x = 4.

Thus the co ordinate is (4,0).

Let me know if this is the correct answer.

Answered by pulakmath007
1

SOLUTION

TO DETERMINE

Consider the line L with equation y + 2x = 3. The line L1

is parallel to L and passes through the point (6, –4).

(a) Find the gradient of L1

(b) Find the equation of L1 in the form y = mx + b.

(c) Find the x-coordinate of the point where line L1 crosses the x-axis.

EVALUATION

The given equation of the line L is

y + 2x = 3

Since the line L1 is parallel to L

Let the equation of the line is

y + 2x = c - - - - - - (1)

The line given by Equation 1 passes through the point (6, –4)

So we have

- 4 + ( 2 × 6 ) = c

⇒ - 4 + 12 = c

⇒ c = 8

The equation of the line L1 is

y + 2x = 8

(a) The equation of the line L1 is y + 2x = 8

Which can be rewritten as

y = - 2x + 8

Which is of the form y = mx + b

m = Gradient of the line = - 2

b = y - intercept of the line = 8

The gradient of L1 = - 2

(b) The equation of the line L1 is y + 2x = 8

Which can be rewritten as

y = - 2x + 8

Which is of the form y = mx + b

m = Gradient of the line = - 2

b = y - intercept of the line = 8

(c) The equation of the line L1 is y + 2x = 8

The line crosses the x-axis

So we have y = 0

Putting y = 0 in above equation we get

0 + 2x = 8

⇒ 2x = 8

⇒ x = 4

So the point is (4,0)

Hence the required x-coordinate of the point where line L1 crosses the x-axis = 4

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