
The line of equation x=2 is parallel to the y axis.
a) true
b) false
Answer
436.2k+ views
Hint: A straight line equation has x and y terms. If the equation of the line fulfils the point P(x,y), then the point P is on the line L. The equation for horizontal or parallel lines to the X-axis is y = a, where a is the y – coordinate of the line's points. A straight line that is vertical or parallel to the Y-axis has the equation x = a, where a is the x-coordinate of the points on the line.
Complete step by step answer:
We have to draw the line whose equation is x=2,
We know that the general equation of a straight line is \[y = mx + c\] .
and we know that A straight line that is vertical or parallel to the Y-axis has the equation x = a, where a is the x-coordinate of the points on the line.
And in our question x=2,
so the graph for x=2 is given below.
So, the correct answer is “Option a”.
Note:
A line is a geometry object that is defined as a zero-width object that extends on both sides. A straight line is one that has no bends. A straight line is a line that stretches to infinity on both sides and has no bends. Equation of a Straight Line is a \[ax + by + c = 0\] . Equation of a Line with 2 Points (Slope Point Form) is \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\] .
Complete step by step answer:
We have to draw the line whose equation is x=2,
We know that the general equation of a straight line is \[y = mx + c\] .
and we know that A straight line that is vertical or parallel to the Y-axis has the equation x = a, where a is the x-coordinate of the points on the line.
And in our question x=2,
so the graph for x=2 is given below.

So, the correct answer is “Option a”.
Note:
A line is a geometry object that is defined as a zero-width object that extends on both sides. A straight line is one that has no bends. A straight line is a line that stretches to infinity on both sides and has no bends. Equation of a Straight Line is a \[ax + by + c = 0\] . Equation of a Line with 2 Points (Slope Point Form) is \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\] .
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