How do you work out what the equation of the parallel line to my=2x+3 and passes through (0,7) is?

Parallel lines have the same gradient so the coefficient of x must still be 2. We know one set of x and y values so we can substitute in to get 7=0*2+c and rearrange to get c=7 Thus the equation of the parallel line is y=2x+7.

If the line is parallel to 2x + 3, it will also have a gradient of 2, so we know that y = 2x + c. To work out c, we can input (0, 7) since we know that it lies on the line. 7 = 2x + c, but x = 0 here so c must be 7. Therefore y = 2x + 7.

If the line is parallel then the gradient will be the same on all equations therefore y=mx+c and m=2 for the gradient. If x=0 and y=7 at the point the line goes through, substituting these values in gives 7=0+c, rearrange to get c=7. Therefore the line is y=2x+7

First of all we want to understand the terms in the question. 'Equation' means we want to find some epression containing xs and ys. 'Parallel' means to have the same gradient. 'Line' means that our equation will have the form y=mx+c, where m is the gradient and c is the y-intercept. Now to find our equation we have to find m and c. M is easy to find as by the definition of being parallel to y=2x+c it must have the same gradient. So m=2. Now to find c we substitute our known point into the equation for a line. y=mx+c turns into 7=2*0+c so c=7. So our final equation is y=2x+7. Hope this helps.

First of all we want to understand the terms in the question. 'Equation' means we want to find some epression containing xs and ys. 'Parallel' means to have the same gradient. 'Line' means that our equation will have the form y=mx+c, where m is the gradient and c is the y-intercept. Now to find our equation we have to find m and c. M is easy to find as by the definition of being parallel to y=2x+c it must have the same gradient. So m=2. Now to find c we substitute our known point into the equation for a line. y=mx+c turns into 7=2*0+c so c=7. So our final equation is y=2x+7. Hope this helps.

A line that is parallel must have the same gradient which in this case is 2. Using the linear form of an equation we get y=2x+c. Now the last piece of information we need is c, which is known as the y-intercept. The question states the line must pass through the point (0,7), they have given us x and y values that will satisfy the equation. So we simply substitute them in like so: y=2x+c ---> (7)=2(0)+c ---> 7=c. Now we have the last piece of information we needed and so we know the answer is y=2x+7.

So first of all, let’s do a quick resume of all data that we have: • Equation of the line m: y = 2x + 3 • Point P where Line u passes through. We know that the slope or gradient of two Parallel Lines it is exactly the same. Therefore: • Slope of Line m: y = 2x + 3. Slope = 2 • Slope of Line u: y = 2x + ....... Now that we have this information, we can work out how to use the Point where it passes through: P(x1:y1) ————> P(0;7) So x1 = 0 and y1 = 7. The formula to calculate “the Line Equation” given its Slope and a Point where it passes through is shown below: (y - y1) = m (x - x1) By substituting with the data values: (y-7) = 2 (x - 0) If you solve it, then: y - 7 = 2x ——-> y = 2x +7 -7 when passes on the other side of the “=“ symbol, change the sign of it: -7 )—> +7 Here we got the Equation of “u” Line passing through P and Parallel to “m” Line. u: y = 2x + 3. Line u m: y = 2x + 7. Line m

A line that is parallel has the same gradient. Plug this into y=mx+c (the equation for a straight line) and set y = 7 and x =0 (from the coordinates) in order to find c.

If a line is parallel, then the equation of the parallel line will have the same gradient as the original line. In this case, the gradient of then parallel line will therefore also be 2. So, you start writing y =2x + c. The y intercept however can only be determined by substituting the coordinates into this equation. In this case, x = 0 and y = 7. So you write 7 = (2*0) + c. Therefore c = 7 The equation of the parallel line is therefore y = 2x + 7