# 如何代数求两条线的交点

## 目录:

- and y{displaystyle y}

-coordinates. Because both lines pass through that point, you know that the x{displaystyle x}

- and y{displaystyle y}

- coordinates must satisfy both equations. With a couple extra techniques, you can find the intersections of parabolas and other quadratic curves using similar logic.

## Steps

### Method 1 of 2: Finding the Intersection of Two Straight Lines

on the left side.

If necessary, rearrange the equation so y{displaystyle y}

is alone on one side of the equal sign. If the equation uses f(x){displaystyle f(x)}

or g(x){displaystyle g(x)}

, separate this term instead. Remember, you can cancel out terms by performing the same action to both sides.

• If you do not know the equations, find them based on the information you have.
• Example:

Your two lines are y=x+3{displaystyle y=x+3}

and y−12=−2x{displaystyle y-12=-2x}

. To get y{displaystyle y}

alone in the second equation, add 12 to each side: y=12−2x{displaystyle y=12-2x}

#### 步骤 2. 设置等式的右边彼此相等。

and y{displaystyle y}

values; this is where the lines cross. Both equations have just y{displaystyle y}

on the left side, so we know the right sides are equal to each other. Write a new equation that represents this.

### For example, if you want to know where the lines y = x + 3 crosses y = 12 - 2x, you'd equate them by writing x + 3 = 12 - 2x

#### 步骤 3. 求解 x。

. Solve this using algebra, by performing the same operation on both sides. Get the x{displaystyle x}

terms on one side of the equation, then put it in the form x=??{displaystyle x=??}

. (If this is impossible, skip down to the end of this section.)

• Example:

x+3=12−2x{displaystyle x+3=12-2x}

to each side:

• 3x+3=12{displaystyle 3x+3=12}

• Subtract 3 from each side:
• 3x=9{displaystyle 3x=9}

• Divide each side by 3:
• x=3{displaystyle x=3}

-value to solve for y{displaystyle y}

Choose the equation for either line. Replace every x{displaystyle x}

in the equation with the answer you found. Do the arithmetic to solve for y{displaystyle y}

• Example:

x=3{displaystyle x=3}

and y=x+3{displaystyle y=x+3}

• y=3+3{displaystyle y=3+3}

• y=6{displaystyle y=6}

#### 步骤 5. 检查您的工作。

-value into the other equation and see if you get the same result. If you get a different solution for y{displaystyle y}

, go back and check your work for mistakes.

• Example:

x=3{displaystyle x=3}

and y=12−2x{displaystyle y=12-2x}

• y=12−2(3){displaystyle y=12-2(3)}

• y=12−6{displaystyle y=12-6}

• y=6{displaystyle y=6}
• This is the same answer as before. We did not make any mistakes.

and y{displaystyle y}

coordinates of the intersection.

You've now solved for the x{displaystyle x}

-value and y{displaystyle y}

-value of the point where the two lines intersect. Write down the point as a coordinate pair, with the x{displaystyle x}

-value as the first number.

• Example:

x=3{displaystyle x=3}

and y=6{displaystyle y=6}

• The two lines intersect at (3, 6).

#### 步骤 7. 处理异常结果。

. This doesn't always mean you made a mistake. There are two ways a pair of lines can lead to a special solution:

• If the two lines are parallel, they do not intersect. The x{displaystyle x}

terms will cancel out, and your equation will simplify to a false statement (such as 0=1{displaystyle 0=1} 线不相交 “或者 没有真正的解决方案"作为你的回答。

• 如果这两个方程描述同一条线，则它们处处“相交”。 x{displaystyle x}

terms will cancel out and your equation will simplify to a true statement (such as 3=3{displaystyle 3=3} 两条线是一样的"作为你的回答。

• ### 方法 2 of 2：二次方程问题

#### 步骤 1. 认识二次方程。

在二次方程中，一个或多个变量的平方 (x2{displaystyle x^{2}}

or y2{displaystyle y^{2}}

), and there are no higher powers. The lines these equations represent are curved, so they can intersect a straight line at 0, 1, or 2 points. This section will teach you how to find the 0, 1, or 2 solutions to your problem.

• Expand equations with parentheses to check whether they're quadratics. For example, y=(x+3)(x){displaystyle y=(x+3)(x)}

is quadratic, since it expands into y=x2+3x.{displaystyle y=x^{2}+3x.}

• Equations for a circle or ellipse have both an x2{displaystyle x^{2}}

and a y2{displaystyle y^{2}}

term. If you're having trouble with these special cases, see the Tips section below.

#### 步骤 2. 根据 y 写出方程。

如有必要，重写每个方程，使 y 单独位于一侧。

• 例子：

求 x2+2x−y=−1{displaystyle x^{2}+2x-y=-1} 的交点

and y=x+7{displaystyle y=x+7}

• Rewrite the quadratic equation in terms of y:
• y=x2+2x+1{displaystyle y=x^{2}+2x+1}
• and y=x+7{displaystyle y=x+7}

.
• This example has one quadratic equation and one linear equation. Problems with two quadratic equations are solved in a similar way.

#### 步骤 3. 组合两个方程以抵消 y。

一旦您将两个方程都设置为等于 y，您就知道没有 y 的两条边彼此相等。

• 例子：

y=x2+2x+1{displaystyle y=x^{2}+2x+1}

and y=x+7{displaystyle y=x+7}

• x2+2x+1=x+7{displaystyle x^{2}+2x+1=x+7}

#### 步骤 4. 排列新方程，使一侧为零。

使用标准代数技术获得一侧的所有项。这将设置问题，以便我们可以在下一步中解决它。

• 例子：

x2+2x+1=x+7{displaystyle x^{2}+2x+1=x+7}

• Subtract x from each side:
• x2+x+1=7{displaystyle x^{2}+x+1=7}

• Subtract 7 from each side:
• x2+x−6=0{displaystyle x^{2}+x-6=0}

#### 步骤 5. 求解二次方程。

将一侧设置为零后，可以使用三种方法来求解二次方程。不同的人发现不同的方法更容易。您可以阅读有关二次公式或“完成平方”的内容，或者按照这个因式分解方法的示例进行操作：

• 例子：

x2+x−6=0{displaystyle x^{2}+x-6=0}

• The goal of factoring is to find the two factors that multiply together to make this equation. Starting with the first term, we know x2{displaystyle x^{2}}

can divide into x, and x. Write down (x)(x) = 0 to show this.

• The last term is -6. List each pair of factors that multiply to make negative six: −6∗1{displaystyle -6*1}

, −3∗2{displaystyle -3*2}

, −2∗3{displaystyle -2*3}

, and −1∗6{displaystyle -1*6}

• The middle term is x (which you could write as 1x). Add each pair of factors together until you get 1 as an answer. The correct pair of factors is −2∗3{displaystyle -2*3}

, since −2+3=1{displaystyle -2+3=1}

• Fill out the gaps in your answer with this pair of factors: (x−2)(x+3)=0{displaystyle (x-2)(x+3)=0}
• .

#### 步骤 6. 留意 x 的两个解。

如果您工作得太快，您可能会找到解决问题的一种方法，而没有意识到还有第二种方法。以下是如何找到在两点相交的线的两个 x 值：

• 例子 （因式分解）：我们最终得到方程 (x−2)(x+3)=0{displaystyle (x-2)(x+3)=0}

. If either of the factors in parentheses equal 0, the equation is true. One solution is x−2=0{displaystyle x-2=0}

→ x=2{displaystyle x=2}

. The other solution is x+3=0{displaystyle x+3=0}

→ x=−3{displaystyle x=-3}

.
• Example (quadratic equation or complete the square): If you used one of these methods to solve your equation, a square root will show up. For example, our equation becomes x=(−1+25)/2{displaystyle x=(-1+{sqrt {25}})/2}
• . Remember that a square root can simplify to two different solutions: 25=5∗5{displaystyle {sqrt {25}}=5*5}

, and 25=(−5)∗(−5){displaystyle {sqrt {25}}=(-5)*(-5)}

. Write two equations, one for each possibility, and solve for x in each one.

#### 步骤 7. 用一个或零个解决方案解决问题。

两条几乎不接触的线只有一个交点，两条从不接触的线的交点为零。以下是识别这些的方法：

• 一种解决方案：将问题分解为两个相同的因子 ((x-1)(x-1) = 0)。当代入二次公式时，平方根项是 0{displaystyle {sqrt {0}}}

. You only need to solve one equation.

• No real solution: There are no factors that satisfy the requirements (summing to the middle term). When plugged into the quadratic formula, you get a negative number under the square root sign (such as −2{displaystyle {sqrt {-2}}}  #### 步骤 8. 将您的 x 值代入任一原始方程。

获得交叉点的 x 值后，将其重新插入到您开始使用的方程之一中。求解 y 以找到 y 值。如果您有第二个 x 值，也请重复此操作。

• 例子：

我们找到了两个解， x=2{displaystyle x=2}

and x=−3{displaystyle x=-3}

. One of our lines has the equation y=x+7{displaystyle y=x+7}

. Plug in y=2+7{displaystyle y=2+7}

and y=−3+7{displaystyle y=-3+7}

, then solve each equation to find that y=9{displaystyle y=9}

and y=4{displaystyle y=4}

#### 步骤 9. 写出点坐标。

现在用坐标形式写出你的答案，用交点的 x 值和 y 值。如果您有两个答案，请确保将正确的 x 值与每个 y 值匹配。

• 例子：

当我们插入 x=2{displaystyle x=2}

, we got y=9{displaystyle y=9} ## 提示

• 圆或椭圆的方程有 x2{displaystyle x^{2}}

term and a y2{displaystyle y^{2}}

term. to find the intersection of a circle and a straight line, solve for x in the linear equation. substitute the solution for x in the circle equation, and you'll have an easier quadratic equation. these problems can have 0, 1, or 2 solutions, as described in the method above.

• a circle and a parabola (or other quadratic) can have 0, 1, 2, 3, or 4 solutions. find the variable that is squared in both equations - let's say it's x2. solve for x2{displaystyle x^{2}}
• and substitute the answer for the x2{displaystyle x^{2}}

in the other equation. solve for y to get 0, 1, or 2 solutions. plug each solution back into the original quadratic equation and solve for x. each of these can have 0, 1, or 2 solutions.