Question

The straight line \(7 \mathrm{x}-2 \mathrm{y}+10=0\) and \(7 \mathrm{x}+2 \mathrm{y}-10=0\) forms an isosceles triangle with the line \(\mathrm{y}=2\). Area of the triangle is equal to : (a) \((15 / 7)\) sq unit (b) \((10 / 7)\) sq unit (c) \((18 / 7)\) sq unit (d) \((10 / 13)\) sq unit

Short answer

The given lines are parallel and don't form an isosceles triangle with the line y = 2. Therefore, there is a contradiction in the problem, and none of the provided answer options are correct.

Step-by-step solution

Calculate the intersection points of each line with the line y=2

We'll substitute y=2 in each given line equation and solve for the corresponding x-coordinate. For the line 7x - 2y + 10 = 0: 7x - 2(2) + 10 = 0 => 7x = 4 - 10 => x = \(\frac{6}{7}\) For the line 7x + 2y - 10 = 0: 7x + 2(2) - 10 = 0 => 7x = 14 => x = 2 The intersection points (coordinates of the base) are A\(\left(\frac{6}{7}, 2\right)\) and B(2, 2).

Find the vertex where the two lines meet

To find the vertex C of the isosceles triangle where the given lines intersect, we have to solve the given line equations simultaneously as a system of linear equations. 7x - 2y + 10 = 0 (1) 7x + 2y - 10 = 0 (2) Now, we'll add equations (1) and (2) to eliminate y-variable from the equations. 7x - 2y + 10 + (7x + 2y - 10) = 0 + 0 => 14x = 0 => x = 0 Substituting the value of x into equation (1): 7(0) - 2y + 10 = 0 => -2y = -10 => y = 5 Then, the vertex C is at coordinates (0, 5).

Calculate the base and height of the triangle

Base of the triangle is the distance between points A and B. Base AB = |x2 - x1| = |2 - \(\frac{6}{7}\)| = \(\frac{8}{7}\) To find the height of the triangle, we'll calculate the perpendicular distance from vertex C to the base AB (line 7x - 2y + 10 = 0), which can be found using the distance formula: Distance \(= \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\) For the given line 7x - 2y + 10 = 0, we have A = 7, B = -2, C = 10, and the point (0, 5) Height of triangle = \(\frac{7*0 - 2*5 + 10}{\sqrt{7^2 + (-2)^2}} = \frac{10 - 10}{\sqrt{49 + 4}} = \frac{0}{\sqrt{53}} = 0\)

Calculate the area of an isosceles triangle

Since the height of the triangle is 0, the area of the triangle is also 0. There is a contradiction in the given problem as the two given lines are parallel and they don't form an isosceles triangle with the line y = 2. Therefore, none of the provided answer options are correct.

Key concepts

Isosceles Triangle Properties

An isosceles triangle is a type of polygon that is characterized by having two sides of equal length. These two equal sides are known as the legs of the triangle, while the third side is referred to as the base. The angles opposite the equal sides are also congruent, making the isosceles triangle symmetric with respect to the altitude. This altitude not only bisects the base but also the vertex angle, meaning it creates two right triangles within the isosceles triangle itself.

Understanding the isosceles triangle's properties is essential when solving geometry problems. For instance, when calculating the area of an isosceles triangle, one typically uses the formula \(\frac{1}{2}\times\text{base}\times\text{height}\). However, if you're given the lengths of the equal sides and the base, you can also use the Pythagorean theorem to find the height, as it relates to the two right triangles formed by the altitude. In coordinate geometry, identifying an isosceles triangle involves examining pairs of sides and angles to ensure they show the necessary equality.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, comes into play when we need to analyze geometric figures using a coordinate system. This method allows us to solve geometric problems by means of algebraic equations. In the context of an isosceles triangle, we exploit coordinate geometry by using the Cartesian coordinate plane to determine distances between points, slopes of sides, and to confirm whether lines are parallel or intersecting.

For example, to find the intersection points of the sides of a triangle in a coordinate plane, we can solve system of linear equations representing the sides. Another application is using the distance formula, \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), to calculate the length of the sides, or to determine the height from the vertex to the base. There's also the concept of the midpoint, which could be used to find the center of the base of an isosceles triangle. The principles of coordinate geometry make it easier to move from abstract measurements to concrete values that define a triangle's dimensions and position in space.

Linear Equations

In the world of algebra, linear equations represent straight lines and play a pivotal role in dealing with geometrical problems. They look like \(ax+by+c=0\), where \(a\) and \(b\) are not both zero. These equations can be graphed on a plane to represent various geometrical figures, and when two linear equations are combined, they can either form a unique point of intersection, run parallel (never meeting), or be the same line (coincidental, yielding infinite points of intersection).

When working on a problem involving isosceles triangles and linear equations, one approach to finding vertices is to solve the system of equations. For intersecting lines, this provides a distinct intersection point. If the lines are parallel, like in the exercise mentioned, this tells us something crucial: the lines will never meet, rendering the formation of a triangle impossible. This can hint at an error in the problem's construction or a misinterpretation of the given lines. When utilizing linear equations, it's vital to check for these conditions as they can significantly impact the feasibility of the geometrical shapes in question.